i wrote a function that receives a string as a char array and converts it to an int:
int makeNumFromString(char Str[])
{
int num = 0, len = 0;
int p;
len = strlen(Str);
for (p = 0; p<len; p++)
{
num = num * 10 + (Str[p] - 48);
}
return num;
}
the problem is that no matter how long the string i input is, when "p" gets to 10 the value of "num" turns to garbage!!!
i tried debbuging and checking the function outside of the larger code but no success.
what could be the problem and how can i fix it?
THANKS
Perhaps your int can only store 32 bits, so the number cannot be higher than 2,147,483,647.
Try using a type for num with more storage, like long.
Related
How can I convert a string to a double using a method that does not include parse or tryparse? I have the program for converting a string to a long, would it be the same for a double? I am a complete newbie.
Code snippets from OP comment below:
public static bool isLong(string s) {
bool n = true;
int a = 0;
s = s.Trim();
for (a = 0; (a < s.Length); a = a + 1) {
n = n && ((s[a] >= '0') && (s[a] <= '9'));
}
return (n);
}
public static long toLong(string s) {
long ret = 0;
int a;
s = s.Trim();
if (isLong(s)) {
for (a = 0; (a< s.Length); a = a + 1) {
ret = (ret * 10) + (s[i] - '0');
}
} else {
}
return (ret);
}
I think I now understand the question. If so, the answer is yes, sort of.
long is an integer type, so processing one digit at a time is fairly straight forward.
double is a floating decimal type, so you have to figure out a way to deal with the decimal period in the middle.
Is this a class assignment or something where you absolutely must write this code on your own? If not, please consider using the library functions that already exist for this purpose, such as stod: http://www.cplusplus.com/reference/string/stod/
I' am trying to convert string to int(like Integer.parseInt() in java) in arduino in order to make some operation's on the numbers. Unfortunately none of my solution's worked.
Until now I tried:
Create char Array and call atoi function:
String StringPassword;
uint8_t *hash;
//Here I define hash
int j;
for (j = 0; j < 20; ++j) {
StringPassword.concat(hash[j]);
}
//Checking String Size
Serial.println("Size");
//Checking String
Serial.println(StringPassword.length());
Serial.println(StringPassword);
int jj;
char PasswordCharArray[StringPassword.length()];
StringPassword.toCharArray(PasswordCharArray, StringPassword.length());
awa = atoi(PasswordCharArray);
Serial.println(awa);
Output:
Size
48
168179819314217391617011617743249832108225513297
18209
Create char Array for null terminated string and call atoi function:
String StringPassword;
uint8_t *hash;
//Here I define hash
int j;
for (j = 0; j < 20; ++j) {
StringPassword.concat(hash[j]);
}
//Checking String Size
Serial.println("Size");
//Checking String
Serial.println(StringPassword.length());
Serial.println(StringPassword);
int jj;
char PasswordCharArray[StringPassword.length()+1];
StringPassword.toCharArray(PasswordCharArray,StringPassword.length()+1);
awa = atoi(PasswordCharArray);
Serial.println(awa);
Output:
Size
48
168179819314217391617011617743249832108225513297
-14511
use toInt Function:
String StringPassword;
uint8_t *hash;
//Here I define hash
int j;
for (j = 0; j < 20; ++j) {
StringPassword.concat(hash[j]);
}
//Checking String Size
Serial.println("Size");
//Checking String
Serial.println(StringPassword.length());
Serial.println(StringPassword);
awa = StringPassword.toInt();
Serial.println(awa);
Output:
Size
48
168179819314217391617011617743249832108225513297
-14511
What is the proper way of changing String to Int so:
awa = 168179819314217391617011617743249832108225513297 ?
And could someone explain to me why my solution's didn't worked? I tried to use the function's that were mentioned on Stackoverflow and Arduino forum to solve this.
The number 168179819314217391617011617743249832108225513297 reaches the maximum integer value limit so therefore this will not convert into an integer.
Try using atol() instead of atoi(). Long numbers can hold more data like the number shown above.
So I have a vector with n one digit elements. I want to make m digit numbers using these elements. m and n will be given by user.
For example:
If the original vector is {0,1,2,3,4,5,6,7,8}, I want the new vector to be {012,345,678}.
So instead of 9 one digit elements, we now have 3 three-digit elements. Once again, the user gives the value of n and m.
Now I have been able to do this by converting the elements to a string and appending them. But the problem is that the array size needs to be very large (about 3 million long).
When I use my logic for this long value, it takes ages to compile. This is what I have currently.
vector<string> list_generator(int *array,int m, int combinations)
{
string ryan, temp;
stringstream aStream;
vector<string> lexicographic;
while (next_permutation(array, array + m))
{
for (int i = 0; i < m; i = i + m)
{
for (unsigned int j = i; j < m+ 1; j++)
aStream << array[i];
temp = aStream.str();
ryan.append(temp);
aStream.str("");
}
lexicographic.push_back(ryan);
ryan.clear();
}
return (lexicographic);
}
Is there a faster way to do this.
I want to know efficient approach for the New Lottery Game problem.
The Lottery is changing! The Lottery used to have a machine to generate a random winning number. But due to cheating problems, the Lottery has decided to add another machine. The new winning number will be the result of the bitwise-AND operation between the two random numbers generated by the two machines.
To find the bitwise-AND of X and Y, write them both in binary; then a bit in the result in binary has a 1 if the corresponding bits of X and Y were both 1, and a 0 otherwise. In most programming languages, the bitwise-AND of X and Y is written X&Y.
For example:
The old machine generates the number 7 = 0111.
The new machine generates the number 11 = 1011.
The winning number will be (7 AND 11) = (0111 AND 1011) = 0011 = 3.
With this measure, the Lottery expects to reduce the cases of fraudulent claims, but unfortunately an employee from the Lottery company has leaked the following information: the old machine will always generate a non-negative integer less than A and the new one will always generate a non-negative integer less than B.
Catalina wants to win this lottery and to give it a try she decided to buy all non-negative integers less than K.
Given A, B and K, Catalina would like to know in how many different ways the machines can generate a pair of numbers that will make her a winner.
For small input we can check all possible pairs but how to do it with large inputs. I guess we represent the binary number into string first and then check permutations which would give answer less than K. But I can't seem to figure out how to calculate possible permutations of 2 binary strings.
I used a general DP technique that I described in a lot of detail in another answer.
We want to count the pairs (a, b) such that a < A, b < B and a & b < K.
The first step is to convert the numbers to binary and to pad them to the same size by adding leading zeroes. I just padded them to a fixed size of 40. The idea is to build up the valid a and b bit by bit.
Let f(i, loA, loB, loK) be the number of valid suffix pairs of a and b of size 40 - i. If loA is true, it means that the prefix up to i is already strictly smaller than the corresponding prefix of A. In that case there is no restriction on the next possible bit for a. If loA ist false, A[i] is an upper bound on the next bit we can place at the end of the current prefix. loB and loK have an analogous meaning.
Now we have the following transition:
long long f(int i, bool loA, bool loB, bool loK) {
// TODO add memoization
if (i == 40)
return loA && loB && loK;
int hiA = loA ? 1: A[i]-'0'; // upper bound on the next bit in a
int hiB = loB ? 1: B[i]-'0'; // upper bound on the next bit in b
int hiK = loK ? 1: K[i]-'0'; // upper bound on the next bit in a & b
long long res = 0;
for (int a = 0; a <= hiA; ++a)
for (int b = 0; b <= hiB; ++b) {
int k = a & b;
if (k > hiK) continue;
res += f(i+1, loA || a < A[i]-'0',
loB || b < B[i]-'0',
loK || k < K[i]-'0');
}
return res;
}
The result is f(0, false, false, false).
The runtime is O(max(log A, log B)) if memoization is added to ensure that every subproblem is only solved once.
What I did was just to identify when the answer is A * B.
Otherwise, just brute force the rest, this code passed the large input.
// for each test cases
long count = 0;
if ((K > A) || (K > B)) {
count = A * B;
continue; // print count and go to the next test case
}
count = A * B - (A-K) * (B-K);
for (int i = K; i < A; i++) {
for (int j = K; j < B; j++) {
if ((i&j) < K) count++;
}
}
I hope this helps!
just as Niklas B. said.
the whole answer is.
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
#define MAX_SIZE 32
int A, B, K;
int arr_a[MAX_SIZE];
int arr_b[MAX_SIZE];
int arr_k[MAX_SIZE];
bool flag [MAX_SIZE][2][2][2];
long long matrix[MAX_SIZE][2][2][2];
long long
get_result();
int main(int argc, char *argv[])
{
int case_amount = 0;
cin >> case_amount;
for (int i = 0; i < case_amount; ++i)
{
const long long result = get_result();
cout << "Case #" << 1 + i << ": " << result << endl;
}
return 0;
}
long long
dp(const int h,
const bool can_A_choose_1,
const bool can_B_choose_1,
const bool can_K_choose_1)
{
if (MAX_SIZE == h)
return can_A_choose_1 && can_B_choose_1 && can_K_choose_1;
if (flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1])
return matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1];
int cnt_A_max = arr_a[h];
int cnt_B_max = arr_b[h];
int cnt_K_max = arr_k[h];
if (can_A_choose_1)
cnt_A_max = 1;
if (can_B_choose_1)
cnt_B_max = 1;
if (can_K_choose_1)
cnt_K_max = 1;
long long res = 0;
for (int i = 0; i <= cnt_A_max; ++i)
{
for (int j = 0; j <= cnt_B_max; ++j)
{
int k = i & j;
if (k > cnt_K_max)
continue;
res += dp(h + 1,
can_A_choose_1 || (i < cnt_A_max),
can_B_choose_1 || (j < cnt_B_max),
can_K_choose_1 || (k < cnt_K_max));
}
}
flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = true;
matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = res;
return res;
}
long long
get_result()
{
cin >> A >> B >> K;
memset(arr_a, 0, sizeof(arr_a));
memset(arr_b, 0, sizeof(arr_b));
memset(arr_k, 0, sizeof(arr_k));
memset(flag, 0, sizeof(flag));
memset(matrix, 0, sizeof(matrix));
int i = 31;
while (i >= 1)
{
arr_a[i] = A % 2;
A /= 2;
arr_b[i] = B % 2;
B /= 2;
arr_k[i] = K % 2;
K /= 2;
i--;
}
return dp(1, 0, 0, 0);
}
How can I add an integer variable to a string and char* variable? for example:
int a = 5;
string St1 = "Book", St2;
char *Ch1 = "Note", Ch2;
St2 = St1 + a --> Book5
Ch2 = Ch1 + a --> Note5
Thanks
The C++ way of doing this is:
std::stringstream temp;
temp << St1 << a;
std::string St2 = temp.str();
You can also do the same thing with Ch1:
std::stringstream temp;
temp << Ch1 << a;
char* Ch2 = new char[temp.str().length() + 1];
strcpy(Ch2, temp.str().c_str());
for char* you need to create another variable that is long enough for both, for instance. You can 'fix' the length of the output string to remove the chance of overrunning the end of the string. If you do that, be careful to make this large enough to hold the whole number, otherwise you might find that book+50 and book+502 both come out as book+50 (truncation).
Here's how to manually calculate the amount of memory required. This is most efficient but error-prone.
int a = 5;
char* ch1 = "Book";
int intVarSize = 11; // assumes 32-bit integer, in decimal, with possible leading -
int newStringLen = strlen(ch1) + intVarSize + 1; // 1 for the null terminator
char* ch2 = malloc(newStringLen);
if (ch2 == 0) { exit 1; }
snprintf(ch2, intVarSize, "%s%i", ch1, a);
ch2 now contains the combined text.
Alternatively, and slightly less tricky and also prettier (but less efficient) you can also do a 'trial run' of printf to get the required length:
int a = 5;
char* ch1 = "Book";
// do a trial run of snprintf with max length set to zero - this returns the number of bytes printed, but does not include the one byte null terminator (so add 1)
int newStringLen = 1 + snprintf(0, 0, "%s%i", ch1, a);
char* ch2 = malloc(newStringLen);
if (ch2 == 0) { exit 1; }
// do the actual printf with real parameters.
snprintf(ch2, newStringLen, "%s%i", ch1, a);
if your platform includes asprintf, then this is a lot easier, since asprintf automatically allocates the correct amount of memory for your new string.
int a = 5;
char* ch1 = "Book";
char* ch2;
asprintf(ch2, "%s%i", ch1, a);
ch2 now contains the combined text.
c++ is much less fiddly, but I'll leave that to others to describe.
You need to create another string large enough to hold the original string followed by the number (i.e. append the character corresponding to each digit of the number to this new string).
Try this out:
char *tmp = new char [ stelen(original) ];
itoa(integer,intString,10);
output = strcat(tmp,intString);
//use output string
delete [] tmp;