Can't determine why sizeof shows different value - c

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));

Answer 2

sizeof gives the size in bytes, where a byte is the size of a char.
printf("\n%d", sizeof(output));
output is a char*, so on your system char*s are 8 bytes large. You must use strlen to obtain the length of a 0-terminated string.

Answer 3

printf("\n%d", sizeof(output)); prints a size of pointer to char. You platform appears to have pointers that are 8 bytes in size.

Answer 4

First:
second = (char *) malloc ((size2+1)*sizeof(char));
Don't typecast the result of malloc().
Second:
sizeof(char) is going to be 1, a char is 1 byte and sizeof() returns the number of bytes required to hold that data type. So you don't need the *sizeof(char), you need that for larger types like int.
Finally:
printf("\nConcatenation of the strings: %s", output); printf("\n%d", sizeof(output));
output is a char * so this is going to give you the sizeof(char *) on your system.
You said in a previous comment:
I know the differences between strlen and sizeof, but using sizeof(output) I wanted to check if everything is fine with memory allocation
Sounds like what you wanted was to verify that you allocated the correct amount of memory for the two concatenated strings. sizeof() won't give you that information. It's in the low level details at this point, you need to do something OS dependent to try and find that information. (ex. in the linux kernel if you kmalloc() memory you can use ksize() to find out exactly how many bytes you got)