So i'm trying to make the variable "name" equal "slider1" but it gives me
the error
initialization with '{...}' expected for aggregate object
the code:
for (int TpNum = 1; TpNum < 2; TpNum++)
{
char name[8] = ("slider" + TpNum );
Enemy name(5, 5, 'r', name);
}
Arrays are aggregate types, and as such are initialized with an initializer list.
More spurious is your attempt to add char*s. You should use C++ std::strings, but here's how you can accomplish it with C strings.
for (int TpNum = 1; TpNum < 2; ++TpNum)
{
char[8] name;
sprintf(name,"slider%d",TpNum);
Enemy name(5, 5, 'r', name);
}
You'll need to include <stdio.h> to use sprintf.
EDIT
Also note that your loop will only execute once, so you could just say
Enemy name(5, 5, 'r', "slider1");
Related
I keep getting an error around handling duplicate characters in key when checking my code for the substitution problem within pset2 of the cs50 course 2020. My code and further details are below - can anyone please help with this? Thanks
The error message it gives me is
:( handles duplicate characters in key
timed out while waiting for program to exit
When I check my code for duplicate characters it seems to work fine (printing Usage: ./substitution key and ending the program)
Code below
# include <stdio.h>
# include <cs50.h>
# include <string.h>
# include <stdlib.h>
# include <ctype.h>
int main(int argc, string argv[])
{
// Check that only one argument submitted
if (argc == 2)
{
// Check that key contains 26 characters
int keylen = strlen(argv[1]);
if (keylen == 26)
{
// Check that all characters are letters
for (int i = 0; i < keylen; i++)
{
bool lettercheck = isalpha(argv[1][i]);
if (lettercheck == true)
{
// THIS IS CAUSING ERROR - Check that no letters have been repeated - put all in lowercase to do so
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1][i - 1];
if (tolower(currentletter) == tolower(previousletter))
{
printf("Usage: ./substitution key\n");
return 1;
}
}
}
else
{
printf("Usage: ./substitution key\n");
return 1;
}
}
}
else
{
printf("Key must contain 26 characters.\n");
return 1;
}
}
else
{
printf("Usage: ./substitution key\n");
return 1;
}
// Get user input
string input = get_string("plaintext: ");
//Transform input using key
for(int i = 0; i < strlen(input); i++)
{
char currentletter = input[i];
int testlower = islower(currentletter);
int testupper = isupper(currentletter);
if (testupper > 0)
{
int j = input[i] - 65;
input[i] = toupper(argv[1][j]);
}
else if (testlower > 0)
{
int j = input[i] - 97;
input[i] = tolower(argv[1][j]);
}
}
printf("ciphertext: %s\n", input);
}
Edit:
Figured out solution - problem was with the second for loop was iterating against i - 1 times instead of n times
Code should have been
charpreviouslletter = argv[1][n]
instead of
charpreviousletter = argv[1][i - 1]
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1]**[i - 1]**
In this loop-
// THIS IS CAUSING ERROR - Check that no letters have been repeated - put all in lowercase to do so
for (int n = 0; n < i; n++)
{
char currentletter = argv[1][i];
char previousletter = argv[1][i - 1];
if (tolower(currentletter) == tolower(previousletter))
{
printf("Usage: ./substitution key\n");
return 1;
}
}
you're comparing only the current character to the previous character. This doesn't work for strings like abcdefca
Notice how, c and a have duplicates - but they're not right next to their originals and hence your logic won't find these duplicates. Your logic will only work for duplicates that are next to each other such as aabcddef.
Instead, you need to take a note of which characters you've encountered whilst looping through. If you encounter a character that you have already encountered, you know there's a duplicate.
Thankfully, the key is only expected to contain all 26 characters of the alphabet without any duplicates. This means you can simply have an int array of 26 slots - each slot counts the number of appearances of the letter at that index. 0th index stands for 'a', 1st for 'b' and so on.
This way, you can very easily get the index of an alphabetic character using letter - 'a', where letter is the alphabetic character. So if the letter was a, you'd get 0, which is indeed the index of 'a'
Also, you have a nested loop while traversing the key, this nested loop also traverses through the key. Except it does it only up until a certain index, the index being the current index of the outer loop. This seems wasteful and weird. Why not simply loop through once, check if current character is an alphabetic letter and also check if this letter has been encountered before. That's all you have to do!
int letter_presence[26];
char upperletter;
string key = argv[1];
if (strlen(key) == KEY_LEN)
{
for (int index = 0; index < KEY_LEN; index++)
{
if (!isalpha(key[index]))
{
// Wrong key - invalid character
printf("Usage: ./substitution key\n");
return 1;
}
if (letter_presence[tolower(key[index]) - 'a'] == 0)
{
// This letter has not been encountered before
letter_presence[upperletter - 'A'] = 1;
}
else
{
// Wrong key - Duplicate letters
return 1;
}
}
// All good
}
I have been working on an exercise from google's dev tech guide. It is called Compression and Decompression you can check the following link to get the description of the problem Challenge Description.
Here is my code for the solution:
public static String decompressV2 (String string, int start, int times) {
String result = "";
for (int i = 0; i < times; i++) {
inner:
{
for (int j = start; j < string.length(); j++) {
if (isNumeric(string.substring(j, j + 1))) {
String num = string.substring(j, j + 1);
int times2 = Integer.parseInt(num);
String temp = decompressV2(string, j + 2, times2);
result = result + temp;
int next_j = find_next(string, j + 2);
j = next_j;
continue;
}
if (string.substring(j, j + 1).equals("]")) { // Si es un bracket cerrado
break inner;
}
result = result + string.substring(j,j+1);
}
}
}
return result;
}
public static int find_next(String string, int start) {
int count = 0;
for (int i = start; i < string.length(); i++) {
if (string.substring(i, i+1).equals("[")) {
count= count + 1;
}
if (string.substring(i, i +1).equals("]") && count> 0) {
count = count- 1;
continue;
}
if (string.substring(i, i +1).equals("]") && count== 0) {
return i;
}
}
return -111111;
}
I will explain a little bit about the inner workings of my approach. It is a basic solution involves use of simple recursion and loops.
So, let's start from the beggining with a simple decompression:
DevTech.decompressV2("2[3[a]b]", 0, 1);
As you can see, the 0 indicates that it has to iterate over the string at index 0, and the 1 indicates that the string has to be evaluated only once: 1[ 2[3[a]b] ]
The core here is that everytime you encounter a number you call the algorithm again(recursively) and continue where the string insides its brackets ends, that's the find_next function for.
When it finds a close brackets, the inner loop breaks, that's the way I choose to make the stop sign.
I think that would be the main idea behind the algorithm, if you read the code closely you'll get the full picture.
So here are some of my concerns about the way I've written the solution:
I could not find a more clean solution to tell the algorithm were to go next if it finds a number. So I kind of hardcoded it with the find_next function. Is there a way to do this more clean inside the decompress func ?
About performance, It wastes a lot of time by doing the same thing again, when you have a number bigger than 1 at the begging of a bracket.
I am relatively to programming so maybe this code also needs an improvement not in the idea, but in the ways It's written. So would be very grateful to get some suggestions.
This is the approach I figure out but I am sure there are a couple more, I could not think of anyone but It would be great if you could tell your ideas.
In the description it tells you some things that you should be awared of when developing the solutions. They are: handling non-repeated strings, handling repetitions inside, not doing the same job twice, not copying too much. Are these covered by my approach ?
And the last point It's about tets cases, I know that confidence is very important when developing solutions, and the best way to give confidence to an algorithm is test cases. I tried a few and they all worked as expected. But what techniques do you recommend for developing test cases. Are there any softwares?
So that would be all guys, I am new to the community so I am open to suggestions about the how to improve the quality of the question. Cheers!
Your solution involves a lot of string copying that really slows it down. Instead of returning strings that you concatenate, you should pass a StringBuilder into every call and append substrings onto that.
That means you can use your return value to indicate the position to continue scanning from.
You're also parsing repeated parts of the source string more than once.
My solution looks like this:
public static String decompress(String src)
{
StringBuilder dest = new StringBuilder();
_decomp2(dest, src, 0);
return dest.toString();
}
private static int _decomp2(StringBuilder dest, String src, int pos)
{
int num=0;
while(pos < src.length()) {
char c = src.charAt(pos++);
if (c == ']') {
break;
}
if (c>='0' && c<='9') {
num = num*10 + (c-'0');
} else if (c=='[') {
int startlen = dest.length();
pos = _decomp2(dest, src, pos);
if (num<1) {
// 0 repetitions -- delete it
dest.setLength(startlen);
} else {
// copy output num-1 times
int copyEnd = startlen + (num-1) * (dest.length()-startlen);
for (int i=startlen; i<copyEnd; ++i) {
dest.append(dest.charAt(i));
}
}
num=0;
} else {
// regular char
dest.append(c);
num=0;
}
}
return pos;
}
I would try to return a tuple that also contains the next index where decompression should continue from. Then we can have a recursion that concatenates the current part with the rest of the block in the current recursion depth.
Here's JavaScript code. It takes some thought to encapsulate the order of operations that reflects the rules.
function f(s, i=0){
if (i == s.length)
return ['', i];
// We might start with a multiplier
let m = '';
while (!isNaN(s[i]))
m = m + s[i++];
// If we have a multiplier, we'll
// also have a nested expression
if (s[i] == '['){
let result = '';
const [word, nextIdx] = f(s, i + 1);
for (let j=0; j<Number(m); j++)
result = result + word;
const [rest, end] = f(s, nextIdx);
return [result + rest, end]
}
// Otherwise, we may have a word,
let word = '';
while (isNaN(s[i]) && s[i] != ']' && i < s.length)
word = word + s[i++];
// followed by either the end of an expression
// or another multiplier
const [rest, end] = s[i] == ']' ? ['', i + 1] : f(s, i);
return [word + rest, end];
}
var strs = [
'2[3[a]b]',
'10[a]',
'3[abc]4[ab]c',
'2[2[a]g2[r]]'
];
for (const s of strs){
console.log(s);
console.log(JSON.stringify(f(s)));
console.log('');
}
I have the below string declaration and i am using the string in for loop:
String[] values = new String["A","B","C"]
for (int i = 0, length = values.length; i < length; i++)
{
getData(values[i], i, length);
}
Throwing an error "unexpected token: = #" at for loop line.
Groovy is different for creating string arrays, you'd do
String[] values = ['A', 'B', 'C']
You could also do what you're trying to do with
def values = ["A","B","C"]
values.eachWithIndex { item, idx ->
getData(item, idx, values.size())
}
If you want to write valid Java, you have to initialize your array with the values wanted in curly braces and that looks like:
String[] values = new String[]{"A","B","C"};
for (int i = 0, length = values.length; i < length; i++) {
getData(values[i], i, length);
}
If you want to do that in Groovy, just use:
String[] values = [ 'A', 'B', 'C' ]
values.eachWithIndex { v, i ->
getData(v, i, values.size())
}
In your code example you did not use a semicolon after the initialization of the for loop.
for example,
for(var i= 0; i
compiling without the semicolon will give you an error, not initializing the I variable.
Code shown below is intended to firstly manipulate the content of string p and then print its content. I have tested the code using two while loops and a printf statement.
#include <stdio.h>
#include <string.h>
int main(void) {
char p[10] = "Hello";
char *a = p;
while(*a++)
;
*a++ = '1';
*a++ = '\0';
a = p;
int i;
for(i = 0; i < 10; i++)
putchar(a[i]);
putchar('\n');
for(i = 0; i < 10; i++)
putchar(p[i]);
putchar('\n');
printf("%s\n", p);
return 0;
}
Output:
xxx#ubuntu:~$ gcc -o test test.c
xxx#ubuntu:~$ ./test
Hello1
Hello1
Hello
Now, I have two question in my mind, first, Where is all the garbage values in first two lines to be shown after printing those 7 characters (including NULL). Secondly, Why the output of printf statement is 'Hello' while is should be 'Hello1'??
I. There's no garbage, for two reasons. One is that
char p[10] = "Hello";
is equivalent with
char p[10] = { 'H', 'e', 'l', 'l', 'o', 0, 0, 0, 0, 0 };
i. e. it initializes the non-listed members to zero (more precisely, it initializes them as if they were static).
Second, even if they were truly uninitialized, you wouldn't have the right to expect garbage values. Reading an uninitialized variable is just undefined behavior, so there could be anything there.
II. Because you end up writing the '1' past the NUL terminator, since your approach of iterating through the string is wrong. So, the for loop with putchar, which iterates over each character will print all of them, including the past-the-NUL '1', but printf() will terminate at the first NUL encountered. TL;DR: instead of
while *(a++);
you should have written
while (*a)
a++;
Try removing ; after the while and do not increase the a twice...
I am new to c++ programming I have to call a function with following arguments.
int Start (int argc, char **argv).
When I try to call the above function with the code below I get run time exceptions. Can some one help me out in resolving the above problem.
char * filename=NULL;
char **Argument1=NULL;
int Argument=0;
int j = 0;
int k = 0;
int i=0;
int Arg()
{
filename = "Globuss -dc bird.jpg\0";
for(i=0;filename[i]!=NULL;i++)
{
if ((const char *)filename[i]!=" ")
{
Argument1[j][k++] = NULL; // Here I get An unhandled
// exception of type
//'System.NullReferenceException'
// occurred
j++;
k=0;
}
else
{
(const char )Argument1[j][k] = filename [j]; // Here I also i get exception
k++;
Argument++;
}
}
Argument ++;
return 0;
}
Start (Argument,Argument1);
Two things:
char **Argument1=NULL;
This is pointer to pointer, You need to allocate it with some space in memory.
*Argument1 = new char[10];
for(i=0, i<10; ++i) Argument[i] = new char();
Don't forget to delete in the same style.
You appear to have no allocated any memory to you arrays, you just have a NULL pointer
char * filename=NULL;
char **Argument1=NULL;
int Argument=0;
int j = 0;
int k = 0;
int i=0;
int Arg()
{
filename = "Globuss -dc bird.jpg\0";
//I dont' know why you have 2D here, you are going to need to allocate
//sizes for both parts of the 2D array
**Argument1 = new char *[TotalFileNames];
for(int x = 0; x < TotalFileNames; x++)
Argument1[x] = new char[SIZE_OF_WHAT_YOU_NEED];
for(i=0;filename[i]!=NULL;i++)
{
if ((const char *)filename[i]!=" ")
{
Argument1[j][k++] = NULL; // Here I get An unhandled
// exception of type
//'System.NullReferenceException'
// occurred
j++;
k=0;
}
else
{
(const char )Argument1[j][k] = filename [j]; // Here I also i get exception
k++;
Argument++;
}
}
Argument ++;
return 0;
}
The first thing you have to do is to find the number of the strings you will have. Thats easy done with something like:
int len = strlen(filename);
int numwords = 1;
for(i = 0; i < len; i++) {
if(filename[i] == ' ') {
numwords++;
// eating up all spaces to not count following ' '
// dont checking if i exceeds len, because it will auto-stop at '\0'
while(filename[i] == ' ') i++;
}
}
In the above code i assume there will be at least one word in the filename (i.e. it wont be an empty string).
Now you can allocate memory for Argument1.
Argument1 = new char *[numwords];
After that you have two options:
use strtok (http://www.cplusplus.com/reference/clibrary/cstring/strtok/)
implement your function to split a string
That can be done like this:
int i,cur,last;
for(i = last = cur = 0; cur < len; cur++) {
while(filename[last] == ' ') { // last should never be ' '
last++;
}
if(filename[cur] == ' ') {
if(last < cur) {
Argument1[i] = new char[cur-last+1]; // +1 for string termination '\0'
strncpy(Argument1[i], &filename[last], cur-last);
last = cur;
}
}
}
The above code is not optimized, i just tried to make it as easy as possible to understand.
I also did not test it, but it should work. Assumptions i made:
string is null terminated
there is at least 1 word in the string.
Also whenever im referring to a string, i mean a char array :P
Some mistakes i noticed in your code:
in c/c++ " " is a pointer to a const char array which contains a space.
If you compare it with another " " you will compare the pointers to them. They may (and probably will) be different. Use strcmp (http://www.cplusplus.com/reference/clibrary/cstring/strcmp/) for that.
You should learn how to allocate dynamically memory. In c you can do it with malloc, in c++ with malloc and new (better use new instead of malloc).
Hope i helped!
PS if there is an error in my code tell me and ill fix it.