In my MPI code in C, i'm receiving a word from each of my slave processes. I want to add all these words to an char array in master side (part of code below). I can print these words but not collect them into a single char array.
(I consider max word length as 10, and number of slave's as slavenumber)
char* word = (char*)malloc(sizeof(char)*10);
char words[slavenumber*10];
for (int p = 0; p<slavenumber; p++){
MPI_Recv(word, 10, MPI_CHAR, p, 0,MPI_COMM_WORLD, MPI_STATUS_IGNORE);
printf("Word: %s\n", word); //it works fine
words[p*10] = *word; //This does not work, i think there is a problem here.
}
printf(words); //This does not work correctly, it gives something like: ��>;&�>W�
Can anybody help me on this?
Let's break it down line by line
// allocate a buffer large enough to hold 10 elements of type `char`
char* word = (char*)malloc(sizeof(char)*10);
// define a variable-length-array large enough to
// hold 10*slavenumber elements of `char`
char words[slavenumber*10];
for (int p = 0; p<slavenumber; p++){
// dereference `word` which is exactly the same as writing
// `word[0]` assigning it to `words[p*10]`
words[p*10] = *word;
// words[p*10+1] to words[p*10+9] are unchanged,
// i.e. uninitialized
}
// printing from an array. For this to work properly all
// accessed elements must be initialized and the buffer
// terminated by a null byte. You have neither
printf(words);
Because you left elements uninitialized and didn't null terminate, you're invoking undefined behavior. Be happy that you didn't get demons crawl out of your nose.
In seriousness though, in C you can copy strings by mere assignment. Your usage case calls for strncpy.
for (int p = 0; p<slavenumber; p++){
strncpy(&words[p*10], word, 10);
}
Related
I'm trying to read a char value using fgets as follows:
int main() {
char m_cityCharCount;
// Input the number of cities
fgets(&m_cityCharCount, 4, stdin);
return 0;
}
Visual Studio returns this error after the code is executed - Stack around the variable m_cityCharCount was corrupted
Is there something I can do about it?
m_cityCharCount is a char, it can hold one char at the most, but you are telling fgets it is 4 bytes buffer. Even if you input nothing but hit the enter key, fgets will store the new line AND the null terminator to the buffer, which of cause is a serious problem. You need a bigger buffer to do fgets:
char str[4096];
fgets(str, sizeof str, stdin);
First parameter of fgets() is pointer on buffer (size of it should be great or equals than second parameter. But sizeof(char) == 1)
int main() {
char m_cityCharCount[4];
// Input the number of cities
fgets(m_cityCharCount, 4, stdin);
return 0;
}
In a sequence S of n characters; each character may occur many times in the sequence. You want to find the longest subsequence of S where all occurrences of the same character are together in one place;
For ex. if S = aaaccaaaccbccbbbab, then the longest such subsequence(answer) is aaaaaaccccbbbb i.e= aaa__aaacc_ccbbb_b.
In other words, any alphabet character that appears in S may only appear in one contiguous block in the subsequence. If possible, give a polynomial time
algorithm to determine the solution.
Design
Below I give a C++ implementation of a dynamic programming algorithm that solves this problem. An upper bound on the running time (which is probably not tight) is given by O(g*(n^2 + log(g))), where n is the length of the string and g is the number of distinct subsequences in the input. I don't know a good way to characterise this number, but it can be as bad as O(2^n) for a string consisting of n distinct characters, making this algorithm exponential-time in the worst case. It also uses O(ng) space to hold the DP memoisation table. (A subsequence, unlike a substring, may consist of noncontiguous character from the original string.) In practice, the algorithm will be fast whenever the number of distinct characters is small.
The two key ideas used in coming up with this algorithm were:
Every subsequence of a length-n string is either (a) the empty string or (b) a subsequence whose first element is at some position 1 <= i <= n and which is followed by another subsequence on the suffix beginning at position i+1.
If we append characters (or more specifically character positions) one at a time to a subsequence, then in order to build all and only the subsequences that satisfy the validity criteria, whenever we add a character c, if the previous character added, p, was different from c, then it is no longer possible to add any p characters later on.
There are at least 2 ways to manage the second point above. One way is to maintain a set of disallowed characters (e.g. using a 256-bit array), which we add to as we add characters to the current subsequence. Every time we want to add a character to the current subsequence, we first check whether it is allowed.
Another way is to realise that whenever we have to disallow a character from appearing later in the subsequence, we can achieve this by simply deleting all copies of the character from the remaining suffix, and using this (probably shorter) string as the subproblem to solve recursively. This strategy has the advantage of making it more likely that the solver function will be called multiple times with the same string argument, which means more computation can be avoided when the recursion is converted to DP. This is how the code below works.
The recursive function ought to take 2 parameters: the string to work on, and the character most recently appended to the subsequence that the function's output will be appended to. The second parameter must be allowed to take on a special value to indicate that no characters have been appended yet (which happens in the top-level recursive case). One way to accomplish this would be to choose a character that does not appear in the input string, but this introduces a requirement not to use that character. The obvious workaround is to pass a 3rd parameter, a boolean indicating whether or not any characters have already been added. But it's slightly more convenient to use just 2 parameters: a boolean indicating whether any characters have been added yet, and a string. If the boolean is false, then the string is simply the string to be worked on. If it is true, then the first character of the string is taken to be the last character added, and the rest is the string to be worked on. Adopting this approach means the function takes only 2 parameters, which simplifies memoisation.
As I said at the top, this algorithm is exponential-time in the worst case. I can't think of a way to completely avoid this, but some optimisations can help certain cases. One that I've implemented is to always add maximal contiguous blocks of the same character in a single step, since if you add at least one character from such a block, it can never be optimal to add fewer than the entire block. Other branch-and-bound-style optimisations are possible, such as keeping track of a globally best string so far and cutting short the recursion whenever we can be certain that the current subproblem cannot produce a longer one -- e.g. when the number of characters added to the subsequence so far, plus the total number of characters remaining, is less than the length of the best subsequence so far.
Code
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
class RunFinder {
string s;
map<string, string> memo[2]; // DP matrix
// If skip == false, compute the longest valid subsequence of t.
// Otherwise, compute the longest valid subsequence of the string
// consisting of t without its first character, taking that first character
// to be the last character of a preceding subsequence that we will be
// adding to.
string calc(string const& t, bool skip) {
map<string, string>::iterator m(memo[skip].find(t));
// Only calculate if we haven't already solved this case.
if (m == memo[skip].end()) {
// Try the empty subsequence. This is always valid.
string best;
// Try starting a subsequence whose leftmost position is one of
// the remaining characters. Instead of trying each character
// position separately, consider only contiguous blocks of identical
// characters, since if we choose one character from this block there
// is never any harm in choosing all of them.
for (string::const_iterator i = t.begin() + skip; i != t.end();) {
if (t.end() - i < best.size()) {
// We can't possibly find a longer string now.
break;
}
string::const_iterator next = find_if(i + 1, t.end(), bind1st(not_equal_to<char>(), *i));
// Just use next - 1 to cheaply give us an extra char at the start; this is safe
string u(next - 1, t.end());
u[0] = *i; // Record the previous char for the recursive call
if (skip && *i != t[0]) {
// We have added a new segment that is different from the
// previous segment. This means we can no longer use the
// character from the previous segment.
u.erase(remove(u.begin() + 1, u.end(), t[0]), u.end());
}
string v(i, next);
v += calc(u, true);
if (v.size() > best.size()) {
best = v;
}
i = next;
}
m = memo[skip].insert(make_pair(t, best)).first;
}
return (*m).second;
}
public:
RunFinder(string s) : s(s) {}
string calc() {
return calc(s, false);
}
};
int main(int argc, char **argv) {
RunFinder rf(argv[1]);
cout << rf.calc() << '\n';
return 0;
}
Example results
C:\runfinder>stopwatch runfinder aaaccaaaccbccbbbab
aaaaaaccccbbbb
stopwatch: Terminated. Elapsed time: 0ms
stopwatch: Process completed with exit code 0.
C:\runfinder>stopwatch runfinder abbaaasdbasdnfa,mnbmansdbfsbdnamsdnbfabbaaasdbasdnfa,mnbmansdbfsbdnamsdnbfabbaaasdbasdnfa,mnbmansdbfsbdnamsdnbfabbaaasdbasdnfa,mnbmansdbfsbdnamsdnbf
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa,mnnsdbbbf
stopwatch: Terminated. Elapsed time: 609ms
stopwatch: Process completed with exit code 0.
C:\runfinder>stopwatch -v runfinder abcdefghijklmnopqrstuvwxyz123456abcdefghijklmnop
stopwatch: Command to be run: <runfinder abcdefghijklmnopqrstuvwxyz123456abcdefghijklmnop>.
stopwatch: Global memory situation before commencing: Used 2055507968 (49%) of 4128813056 virtual bytes, 1722564608 (80%) of 2145353728 physical bytes.
stopwatch: Process start time: 21/11/2012 02:53:14
abcdefghijklmnopqrstuvwxyz123456
stopwatch: Terminated. Elapsed time: 8062ms, CPU time: 7437ms, User time: 7328ms, Kernel time: 109ms, CPU usage: 92.25%, Page faults: 35473 (+35473), Peak working set size: 145440768, Peak VM usage: 145010688, Quota peak paged pool usage: 11596, Quota peak non paged pool usage: 1256
stopwatch: Process completed with exit code 0.
stopwatch: Process completion time: 21/11/2012 02:53:22
The last run, which took 8s and used 145Mb, shows how it can have problems with strings containing many distinct characters.
EDIT: Added in another optimisation: we now exit the loop that looks for the place to start the subsequence if we can prove that it cannot possibly be better than the best one discovered so far. This drops the time needed for the last example from 32s down to 8s!
EDIT: This solution is wrong for OP's problem. I'm not deleting it because it might be right for someone else. :)
Consider a related problem: find the longest subsequence of S of consecutive occurrences of a given character. This can be solved in linear time:
char c = . . .; // the given character
int start = -1;
int bestStart = -1;
int bestLength = 0;
int currentLength = 0;
for (int i = 0; i < S.length; ++i) {
if (S.charAt(i) == c) {
if (start == -1) {
start = i;
}
++currentLength;
} else {
if (currentLength > bestLength) {
bestStart = start;
bestLength = currentLength;
}
start = -1;
currentLength = 0;
}
}
if (bestStart >= 0) {
// longest sequence of c starts at bestStart
} else {
// character c does not occur in S
}
If the number of distinct characters (call it m) is reasonably small, just apply this algorithm in parallel to each character. This can be easily done by converting start, bestStart, currentLength, bestLength to arrays m long. At the end, scan the bestLength array for the index of the largest entry and use the corresponding entry in the bestStart array as your answer. The total complexity is O(mn).
import java.util.*;
public class LongestSubsequence {
/**
* #param args
*/
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.next();
execute(str);
}
static void execute(String str) {
int[] hash = new int[256];
String ans = "";
for (int i = 0; i < str.length(); i++) {
char temp = str.charAt(i);
hash[temp]++;
}
for (int i = 0; i < hash.length; i++) {
if (hash[i] != 0) {
for (int j = 0; j < hash[i]; j++)
ans += (char) i;
}
}
System.out.println(ans);
}
}
Space: 256 -> O(256), I don't if it's correct to say this way..., cause O(256) I think is O(1)
Time: O(n)
I have defined two typedef structs, and the second has the first as an object:
typedef struct
{
int numFeatures;
float* levelNums;
} Symbol;
typedef struct
{
int numSymbols;
Symbol* symbols;
} Data_Set;
I then defined numFeatures and numSymbols and allocate memory for both symbols and levelNums, then fill levelNums inside a for loop with value of the inner loop index just to verify it is working as expected.
Data_Set lung_cancer;
lung_cancer.numSymbols = 5;
lung_cancer.symbols = (Symbol*)malloc( lung_cancer.numSymbols * sizeof( Symbol ) );
lung_cancer.symbols->numFeatures = 3;
lung_cancer.symbols->levelNums = (float*)malloc( lung_cancer.symbols->numFeatures * sizeof( float ) );
for(int symbol = 0; symbol < lung_cancer.numSymbols; symbol++ )
for( int feature = 0; feature < lung_cancer.symbols->numFeatures; feature++ )
*(lung_cancer.symbols->levelNums + symbol * lung_cancer.symbols->numFeatures + feature ) = feature;
for(int symbol = 0; symbol < lung_cancer.numSymbols; symbol++ )
for( int feature = 0; feature < lung_cancer.symbols->numFeatures; feature++ )
cout << *(lung_cancer.symbols->levelNums + symbol * lung_cancer.symbols->numFeatures + feature ) << endl;
return 0;
When levelNums are int I get what I expect( i.e. 0,1,2,0,1,2,...) but when they are float, only the first 3 are correct and the remaining are very small or very large values, not 0,1,2 like expected. I then have two questions:
When allocating memory for symbols, how does it know how big a Symbol is since I have not yet defined how large levelNums will be yet.
How do I get float values into levelNums correctly.
The reason I am doing it like this is this is a data structure that will be sent to a GPU for GPGPU programming in CUDA and arrays are not recognized. I can only send in a continuous block of memory explicitly and the typedef structs are only there for conveying/defining the memory struture of the data.
A couple thing jump out at meet. For one thing, you only allocated a buffer for levelNums of the first symbol. Similarly, your inner loops always loop over the numFeatures of the first symbol.
You're doing a whole lot of dereferencing of arrays, which is fine in general, but the assignment in particular (inside the first set of loops) looks very strange. It's entirely possible I just don't understand what you're trying to do there, but I think it'd be a lot less confusing if you used some square bracket array accessors.
I am getting an int value from one of the analog pins on my Arduino. How do I concatenate this to a String and then convert the String to a char[]?
It was suggested that I try char msg[] = myString.getChars();, but I am receiving a message that getChars does not exist.
To convert and append an integer, use operator += (or member function concat):
String stringOne = "A long integer: ";
stringOne += 123456789;
To get the string as type char[], use toCharArray():
char charBuf[50];
stringOne.toCharArray(charBuf, 50)
In the example, there is only space for 49 characters (presuming it is terminated by null). You may want to make the size dynamic.
Overhead
The cost of bringing in String (it is not included if not used anywhere in the sketch), is approximately 1212 bytes of program memory (flash) and 48 bytes RAM.
This was measured using Arduino IDE version 1.8.10 (2019-09-13) for an Arduino Leonardo sketch.
Risk
There must be sufficient free RAM available. Otherwise, the result may be lockup/freeze of the application or other strange behaviour (UB).
Just as a reference, below is an example of how to convert between String and char[] with a dynamic length -
// Define
String str = "This is my string";
// Length (with one extra character for the null terminator)
int str_len = str.length() + 1;
// Prepare the character array (the buffer)
char char_array[str_len];
// Copy it over
str.toCharArray(char_array, str_len);
Yes, this is painfully obtuse for something as simple as a type conversion, but somehow it's the easiest way.
You can convert it to char* if you don't need a modifiable string by using:
(char*) yourString.c_str();
This would be very useful when you want to publish a String variable via MQTT in arduino.
None of that stuff worked. Here's a much simpler way .. the label str is the pointer to what IS an array...
String str = String(yourNumber, DEC); // Obviously .. get your int or byte into the string
str = str + '\r' + '\n'; // Add the required carriage return, optional line feed
byte str_len = str.length();
// Get the length of the whole lot .. C will kindly
// place a null at the end of the string which makes
// it by default an array[].
// The [0] element is the highest digit... so we
// have a separate place counter for the array...
byte arrayPointer = 0;
while (str_len)
{
// I was outputting the digits to the TX buffer
if ((UCSR0A & (1<<UDRE0))) // Is the TX buffer empty?
{
UDR0 = str[arrayPointer];
--str_len;
++arrayPointer;
}
}
With all the answers here, I'm surprised no one has brought up using itoa already built in.
It inserts the string representation of the integer into the given pointer.
int a = 4625;
char cStr[5]; // number of digits + 1 for null terminator
itoa(a, cStr, 10); // int value, pointer to string, base number
Or if you're unsure of the length of the string:
int b = 80085;
int len = String(b).length();
char cStr[len + 1]; // String.length() does not include the null terminator
itoa(b, cStr, 10); // or you could use String(b).toCharArray(cStr, len);
I am having a hard time in manipulating strings while writing module for linux. My problem is that I have a int Array[10] with different values in it. I need to produce a string to be able send to the buffer in my_read procedure. If my array is {0,1,112,20,4,0,0,0,0,0}
then my output should be:
0:(0)
1:-(1)
2:-------------------------------------------------------------------------------------------------------(112)
3:--------------------(20)
4:----(4)
5:(0)
6:(0)
7:(0)
8:(0)
9:(0)
when I try to place the above strings in char[] arrays some how weird characters end up there
here is the code
int my_read (char *page, char **start, off_t off, int count, int *eof, void *data)
{
int len;
if (off > 0){
*eof =1;
return 0;
}
/* get process tree */
int task_dep=0; /* depth of a task from INIT*/
get_task_tree(&init_task,task_dep);
char tmp[1024];
char A[ProcPerDepth[0]],B[ProcPerDepth[1]],C[ProcPerDepth[2]],D[ProcPerDepth[3]],E[ProcPerDepth[4]],F[ProcPerDepth[5]],G[ProcPerDepth[6]],H[ProcPerDepth[7]],I[ProcPerDepth[8]],J[ProcPerDepth[9]];
int i=0;
for (i=0;i<1024;i++){ tmp[i]='\0';}
memset(A, '\0', sizeof(A));memset(B, '\0', sizeof(B));memset(C, '\0', sizeof(C));
memset(D, '\0', sizeof(D));memset(E, '\0', sizeof(E));memset(F, '\0', sizeof(F));
memset(G, '\0', sizeof(G));memset(H, '\0', sizeof(H));memset(I, '\0', sizeof(I));memset(J, '\0', sizeof(J));
printk("A:%s\nB:%s\nC:%s\nD:%s\nE:%s\nF:%s\nG:%s\nH:%s\nI:%s\nJ:%s\n",A,B,C,D,E,F,G,H,I,J);
memset(A,'-',sizeof(A));
memset(B,'-',sizeof(B));
memset(C,'-',sizeof(C));
memset(D,'-',sizeof(D));
memset(E,'-',sizeof(E));
memset(F,'-',sizeof(F));
memset(G,'-',sizeof(G));
memset(H,'-',sizeof(H));
memset(I,'-',sizeof(I));
memset(J,'-',sizeof(J));
printk("A:%s\nB:%s\nC:%s\nD:%s\nE:%s\nF:%s\nG:%s\nH:%s\nI:%s\nJ:%\n",A,B,C,D,E,F,G,H,I,J);
len = sprintf(page,"0:%s(%d)\n1:%s(%d)\n2:%s(%d)\n3:%s(%d)\n4:%s(%d)\n5:%s(%d)\n6:%s(%d)\n7:%s(%d)\n8:%s(%d)\n9:%s(%d)\n",A,ProcPerDepth[0],B,ProcPerDepth[1],C,ProcPerDepth[2],D,ProcPerDepth[3],E,ProcPerDepth[4],F,ProcPerDepth[5],G,ProcPerDepth[6],H,ProcPerDepth[7],I,ProcPerDepth[8],J,ProcPerDepth[9]);
return len;
}
it worked out with this:
char s[500];
memset(s,'-',498);
for (i=len=0;i<10;++i){
len+=sprintf(page+len,"%d:%.*s(%d)\n",i,ProcPerDepth[i],s,ProcPerDepth[i]);
}
I wonder if there is an easy flag to multiply string char in sprintf. thanx –
Here are a some issues:
You have entirely filled the A, B, C ... arrays with characters. Then, you pass them to an I/O routine that is expecting null-terminated strings. Because your strings are not null-terminated, printk() will keep printing whatever is in stack memory after your object until it finds a null by luck.
Multi-threaded kernels like Linux have strict and relatively small constraints regarding stack allocations. All instances in the kernel call chain must fit into a specific size or something will be overwritten. You may not get any detection of this error, just some kind of downstream crash as memory corruption leads to a panic or a wedge. Allocating large and variable arrays on a kernel stack is just not a good idea.
If you are going to write the tmp[] array and properly nul-terminate it, there is no reason to also initialize it. But if you were going to initialize it, you could do so with compiler-generated code by just saying: char tmp[1024] = { 0 }; (A partial initialization of an aggregate requires by C99 initialization of the entire aggregate.) A similar observation applies to the other arrays.
How about getting rid of most of those arrays and most of that code and just doing something along the lines of:
for(i = j = 0; i < n; ++i)
j += sprintf(page + j, "...", ...)