Question

I am making a program that concatenates two strings given by user. Everything is fine, but I don’t know why the program shows that the sizeof of the final result is 8-bit long. Doesn’t matter how long the strings are, it always shows 8. I guess, that it is the size of char, but I would like to know why it acts this way. This is the code:

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

char* concatenate(char *fir, char *sec)
{
int firLen = strlen(fir);
int secLen = strlen(sec);
int len = firLen + secLen + 1;
int i = 0,c=0;
int *wsk = &i;

char *result = (char *)malloc(len*sizeof(char));

while (fir[i]!='\0')
{
    result[i]=fir[i];
    (*wsk)++;
}

while (sec[c]!='\0')
{
    result[i]=sec[c];
    (*wsk)++;
    c++;
}

result[len-1] = '\0';
return result;
}

int main(int argc, char **argv)
{
char *first, *second, *output;
int size1, size2;

printf("How long will your first string be: ");
scanf("%d", &size1);

first = (char *) malloc ((1+size1)*sizeof(char));
if (!first)
{
    puts("\nError. Can't allocate memory!");
    abort();
}

printf("How long will your second string be: ");
scanf("%d", &size2);

second = (char *) malloc ((size2+1)*sizeof(char));
if (!second)
{
    puts("\nError. Can't allocate memory!");
    abort();
}

printf("\nPlease, type in the first string: ");
scanf("%s",first);

printf("\nPlease, type in the second string: ");
scanf("%s",second);

output = (char *)malloc((size1+size2+1)*sizeof(char));
output = concatenate(first, second);

printf("\nConcatenation of the strings: %s", output);
printf("\n%d", sizeof(output));

free(first);
free(second);
free(output);

getchar();
return 0;
}

Answer 1

Don’t use sizeof to determine string length, use the strlen function, as you’re doing in other parts of your program.

printf("\nConcatenation of the strings: %s", output);
printf("\n%d", strlen(output));

Use sizeof to determine the size of data types, like char, a struct, an array, etc.

You’ve mentioned that you want to use sizeof to check if everything is fine with memory allocation, but you can’t use sizeof this way on a buffer of memory allocated with malloc: you can only rely on the return value of malloc – which you aren’t checking – to know if your allocation was successful.

You can use sizeof to determine the size of an array:

char myStr[13];
printf("%d\n", sizeof(myStr));

But this only works for arrays, and not buffers allocated using malloc.

Also this line creates a memory leak because you overwrite the pointer in the next line:

output = (char *)malloc((size1+size2+1)*sizeof(char));