C string function with pointers returning null

Question

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

void catTo(char *dest, const char *str1, const char *str2)
{
    dest = (char *) malloc(1 + strlen(str1)+ strlen(str2) );
    strcpy(dest, str1);
    strcat(dest, str2);
}

int main(int argc, char** argv) {
    char *str1 = "abcd";
    char *str2 = "defg";
    char *dest;
    catTo(dest, str1, str2);
    printf("%s",dest);

    return 0;
}

I have been trying to get a simple str copy function to work, however, when I print out dest, I get "(null)". I tried, messing around with putting &/* infront of certain variables, but to no prevail.

void catTo(char *dest, : dest won't be modified outside your function. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Feb 2, 2017 at 14:03
when compiling, always enable all the warnings, then fix those warnings. (for gcc, at a minimum use: -Wall -Wextra -pedantic I also use: -Wconversion -std=gnu99 ) — user3629249
– user3629249, Commented Feb 3, 2017 at 18:26
when calling any of the heap memory allocation functions (malloc, calloc, realloc) 1) do not cast the returned value. The type is already void* so can be assigned to any other pointer. Casting just clutters the code, making it more difficult to understand, debug, maintain. 2) always check (!=NULL) the returned value to assure the operation was successful — user3629249
– user3629249, Commented Feb 3, 2017 at 18:35

Jean-François Fabre · Accepted Answer · 2017-02-02 14:07:21Z

4

don't pass dest as an input parameter. Its start value is not needed (and undefined) and is not modified by the catTo function call, which explains your problem.

Better do this:

char *catTo(const char *str1, const char *str2)
{
    char *dest = malloc(1 + strlen(str1)+ strlen(str2) );
    strcpy(dest, str1);
    strcat(dest, str2);
    return dest;
}

and in the caller:

char *dest = catTo(str1, str2);

edited Feb 2, 2017 at 14:07

answered Feb 2, 2017 at 14:04

Jean-François Fabre♦

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1 Comment

Vittorio Romeo Over a year ago

Did you mean to write char dest = /*...*/in catTo?

Vittorio Romeo · Accepted Answer · 2017-02-02 14:11:27Z

4

I tried, messing around with putting &/* infront of certain variables, but to no prevail.

That's a terrible way of solving problems - you should review your knowledge on pointers and value semantics instead of randomly adding operators to your code.

Regardless, the issue is that you're passing dest as a char* and then assigning to dest inside catTo: since you're copying dest, you're assigning to the local argument, not to the dest instance present in main.

You can solve this issue by passing dest as a char**:

void catTo(char **dest, const char *str1, const char *str2)
{
    *dest = malloc(1 + strlen(str1)+ strlen(str2) );
    strcpy(*dest, str1);
    strcat(*dest, str2);
}

int main(int argc, char** argv) {
    char *str1 = "abcd";
    char *str2 = "defg";
    char *dest;
    catTo(&dest, str1, str2);
    printf("%s",dest);

    return 0;
}

Also, every malloc should be matched by a corresponding free. Don't forget to call free(dest) when you're done with it in main:

int main(int argc, char** argv) {
    char *str1 = "abcd";
    char *str2 = "defg";
    char *dest;
    catTo(&dest, str1, str2);
    printf("%s",dest);
    free(dest);

    return 0;
}

edited Feb 2, 2017 at 14:11

answered Feb 2, 2017 at 14:04

Vittorio Romeo

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2 Comments

Jean-François Fabre Over a year ago

no need for that char* cast in malloc: (char *) malloc => malloc

Quentin Over a year ago

@Jean-FrançoisFabre still not Javascript's level of lenience, but pretty close!

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