CODEWITHSUNDEEP
cstring
Joshua Cook
Joshua Cook

Reputation: 13425

Odd behavior when converting strings to integers in C

Was working on a program taking a mathematical expression as a string and then evaluating it and discovered an odd behavior. Given

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *argv[]){

    int ary[4];
    char * string = "123+11";

    for (int i = 0; i < 3; i++){

        ary[i] = atoi(&string[i]);
        printf("%d\n", ary[i]);
    }

}

I get the output:

123
23
3

Whereas I might have expected I get the output:

1
2
3

Is this part of the atoi() function?

Upvotes: 0

Views: 85

Answers (2)

prince
prince

Reputation: 1149

The answer given by user3758647 is correct. To solve you problem you can use strtok function which tokenizes the input string based on delimiter.

char* string = "123+23+22";
char *token = strtok(string, "+");
int arr[4], i = 0;
arr[i] = atoi(token); //Collect first number here
while (token != NULL)
{
    token = strtok(NULL, " ");
    //You can collect rest of the numbers using atoi function from here
    i++;
    arr[i]  = atoi(token); 
    //Do whatever you want to do with this number here.
}
return 0;

Upvotes: 1

Abhishek Mittal
Abhishek Mittal

Reputation: 366

This is correct behavior because atoi takes a pointer to char as input and convert it into int till it finds "\0" character.

char * string = "123";

"\0" in string is present after 123.

For statement:

ary[0] = atoi(&string[0]);

atoi starts with 1 convert it to int till 123.

For statement:

ary[1] = atoi(&string[1]);

atoi starts with 2 convert it to int till 23.

For statement:

 ary[2] = atoi(&string[2]);

atoi starts with 3 convert it to int till 3.

Please let me know if it is not clear.

Upvotes: 4

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