Returning an Array of Pointers from a Function

Question

I am trying to print the integer value of each character in a given string using a function. When I pass a string to this function, it converts each character in the given string to an integer and then saves it to a result array. I have also created an array of pointers which will hold the address of each character's corresponding integer. But when I return this array of pointers, I am not able to print all integer values correctly.

#include <stdio.h>
int *c2i(char *str); 

int main(){

    char s[20];
    int *p, i;
    printf("Enter your string:\n");
    scanf("%s", s);
    p=c2i(s);
    for(i=0; i<4; i++){
        printf("\n%d\t%d\n", *p, p);
        p++;
    }

    return 0;
}

int *c2i(char *str){

    int i=0, result[20], *ptr[20];

    while(*(str+i)!='\0'){
        result[i]=*(str+i) & 0xff;
        ptr[i]=&result[i];
        i++;
    }
    printf("%d %d %d %d %d %d %d %d", *ptr[0], *ptr[1], *ptr[2], *ptr[3], ptr[0], ptr[1], ptr[2], ptr[3]);

    return *ptr;
}

I have included many printf statements just for debugging purposes. The output of the program when I try to run the code above is:

Enter your string:

abcd

97 98 99 100 6356588 6356592 6356596 6356600

97 6356588

1999382056 6356592

20 6356596

1 6356600

I am only getting the first value correctly when I print it in the main() function. When I increment the pointer and try to print the next values, I am getting garbage results. I can see that the address is the same but the values have been corrupted. What am I doing wrong here?

Accepted Answer

You can't return the stack-allocated array since once the function call finishes, it is no longer marked in use. Instead, you should malloc the array, for example int *ptr = malloc(sizeof(int)*20);. Also remember to free the array when done with it (it's good practice). See these posts for details

Returning an array using C

returning a local variable from function in C

Accepted Answer

#include <stdio.h>

int** c2i(char *str) {

    int i=0, *result, **ptr;
    result = (int *)malloc(20*sizeof(int)) ;
    ptr = (int **)malloc(10*sizeof(int *)) ;

    while(*(str+i)!='\0'){
        result[i]=*(str+i) & 0xff;
        ptr[i]=&result[i];
        i++;
    }
    printf("%d %d %d %d %p %p %p %p", *ptr[0], *ptr[1], *ptr[2], *ptr[3], ptr[0], ptr[1], ptr[2], ptr[3]);

    return ptr;
}
int main(){

    char s[20];
    int **p, i;
    printf("Enter your string:\n");
    scanf("%s", s);
    p=c2i(s);
    for(i=0; i<4; i++){
        printf("\n%d\t%p\n", **p, *p);
        p++;
    }

    return 0;
}

double-pointer might solve your issue, look at your modified code..this gives you the desired result.here i'm attaching the screen shot of execution..

Answer 3

How about this

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

#define MAXLEN 20

void c2i(char *str, int *out, int len); 

int main(){

    char s[MAXLEN];
    int  out[MAXLEN];
    int *p, i;

    printf("Enter your string:\n");
    /* fgets safer that scanf, 
      protects buffer overrun */
    fgets(s, sizeof(s), stdin);
    /* fgets gives you the newline char though, 
      so remove it */
    if(s[strlen(s)-1] == '\n') {
        s[strlen(s)-1] = '\0';
    }

    c2i(s,out,strlen(s));
    for(i=0; i<strlen(s); i++){
        printf("%d\n", out[i]);
    }

    return 0;
}

void c2i(char *str, int *out, int len){

    int i=0;

    for(i=0; i<len; i++)
    {
        out[i]=*(str+i) & 0xff;
    }
}

Passing the out buffer to c2i means you don't need to malloc/free a buffer, but this is probably just a matter of taste.

Using fgets means rather than scanf means you're protected from buffer overruns and your strings can have spaces in them too.

Returning an Array of Pointers from a Function

3 Answers

Must Read

Your Answer

Sarah Jenkins

Returning an Array of Pointers from a Function

3 Answers

Must Read

Your Answer

Sign up or log in

Post as a guest

Sarah Jenkins