Converting a Char Array into an Int array using a Function

Question

I have been tasked with converting a char* array into an int array using a function. I'm new to C so I do apologize if this is a simple fix or an easy solution that I have just failed to notice.

This is the code I currently have:

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

int convert(char* str[], int length);

int main(void)
{
    int result[50], i, length = 4;
    char *str[] = {"7", "1", "14", "15"};

    result[4] = convert(str, 4);
    for(i = 0; i < length; i++) 
    {     
        printf("%d ", result[i]);
    }
    printf("\n");
    return 0;
}

int convert(char* str[], int length)
{
    int i;
    int arr[50];

    for(i = 0; i > length; i++)
    arr[i] = atoi(str[i]);

    return arr[50];
}

The output is currently this:

832 832 832 832

Essentially, what the code is attempting to do is convert

char *str[] = {"7", "1", "14", "15"};

into this (output):

7 1 14 15

Any help would be appreciated! :)

`return arr[50]` that returns a single `int`. So it straight away says you can't possibly get results for multiple ints. Also `arr[50]` is past the end of the array. Perhaps you are trying to return the whole array as a result. See the duplicate post for the correct way to do that. — kaylum, Oct 10 '22 at 05:41
Does this answer your question? [Returning an array using C](https://stackoverflow.com/questions/11656532/returning-an-array-using-c) — kaylum, Oct 10 '22 at 05:42
`for(i = 0; i > length; i++)` is obviously wrong. As is the senseless `return arr[50];`. You have to study arrays before using arrays. — Lundin, Oct 10 '22 at 06:30

Allan Wind · Accepted Answer · 2022-10-10T06:50:08.720

Moved the definition of convert() above above main() so the prototype is not required. As you know the size of the result array it is cleaner to pass it in as an argument, and as @kaylum noted above, the return type of int is clearly incorrect. The loop is wrong (should be i < length). You cannot return a local array arr even if you tried as it will be out of scope when function returns. return arr[50] refers to an element that is out of bounds. In main(), the expression result[4] = sets the 4th element and rest of result contains undefined data.

#include <stdio.h>
#include <stdlib.h>
#define ARRAY_LEN(A) (sizeof(A) / sizeof(*A))

void convert(int length, char *str[length], int result[length]) {
    for(int i = 0; i < length; i++)
        result[i] = atoi(str[i]);
}

int main(void) {
    char *str[] = {"7", "1", "14", "15"};
    int result[ARRAY_LEN(str)];
    convert(ARRAY_LEN(str), str, result);
    for(int i = 0; i < ARRAY_LEN(result); i++) {
        printf("%d ", result[i]);
    }
    printf("\n");
}

Consider using strtol() instead of atoi() so you can implement error handling (atoi() returns 0 on error).

Converting a Char Array into an Int array using a Function

1 Answers1