C - strcat in for loop - c

Question

I m writing a little C program and want to know why my output in the console is "0", "0" [...]? The output i expect is "ab", "ac", [...].
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i;
int j;
char string[] = "abc";
char output[8];
int length = size(&string[0]);
for(i=0; i<length; i++) {
for(j=0; j<length; j++){
char a = string[i];
strcat(output, &a);
char b = string[j];
strcat(output, &b);
printf("%c\n", output);
}
}
return 0;
}

Answer 1

Mistake #1. You have not initialised output[] so strcat() will not validly find a nul terminator to append to.
output[0] = 0;
Mistake #2. strcat() isn't the right way of appending chars anyway.
Mistake #3. Your loop controls aren't right. See below.
Mistake #4. Your length is the size of a char* pointer.
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i, j;
char string[] = "abc";
char output[8];
int length = strlen (string); // corrected
for(i=0; i<length-1; i++) { // amended loop
for(j=i+1; j<length; j++) { // amended loop
output[0] = string [i];
output[1] = string [j];
output[2] = 0; // string terminator
printf("%s\n", output); // uses string type not char
}
}
return 0;
}
Program output:
ab
ac
bc

Answer 2

If I have understood correctly what you are trying to do then the program will look the following way
#include <stdio.h>
int main(int argc, char *argv[])
{
char string[] = "abc";
char output[3];
size_t length = sizeof( string ) - 1;
for ( size_t i = 0; i < length; i++ )
{
for ( size_t j = 0; j < length; j++ )
{
if ( i != j )
{
output[0] = string[i];
output[1] = string[j];
output[2] = '\0';
puts( output );
}
}
}
return 0;
}
The output is
ab
ac
ba
bc
ca
cb
If your compiler does not allow to declare variables within the control statement of the loop then you can declare i and j in the beginning of the program.
size_t i, j;
If you want to include combinations like "aa" then you simply may remove the if statement withing the inner loop.

Answer 3

char a = string[i];
strcat(output, &a);
leads to undefined behavior since strcat expects a null terminated string in the second argument. Same thing applies to:
char b = string[j];
strcat(output, &b);
Perhaps you meant to use:
output[0] = a;
output[1] = b;
output[2] = '\0';
Here's the updated for loop:
for(i=0; i<length; i++) {
for(j=0; j<length; j++){
output[0] = a;
output[1] = b;
output[2] = '\0';
printf("%s\n", output);
// ^^ use %s to print a string, not %c.
}
}

Answer 4

If you want to use strcat you must know that it expects a string not a character and there is an important difference, when you pass &a strcat thinks it is the address of a pointer to a string, and you should get most likely a segmentation fault, here I show your own code, modified to use strcat but you don't really need it for this task.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i;
int j;
char string[] = "abc";
int length = strlen(&string[0]);
for(i = 0 ; i < length ; i++)
{
for(j= i + 1 ; j < length ; j++)
{
/* initialize output to 0 more importantly to have a terminating null byte */
char output[3] = {0};
/*
* create a string and initialize it with 2 char's
* - the first one, the one you want to append to output
* - the second one is required by strcat, to mark the end of the string
*/
char a[2] = {string[i], 0};
strcat(output, a);
/* same as above */
char b[2] = {string[j], 0};
strcat(output, b);
printf("%s\n", output);
}
}
return 0;
}
you could do this without strcat unless you are trying to learn how to use strcat, this is an example of how to do it without strcat.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i;
int j;
char string[] = "abc";
int length = strlen(&string[0]);
for(i = 0 ; i < length ; i++)
{
for(j= i + 1 ; j < length ; j++)
{
char output[3] = {string[i], string[j], 0};
printf("%s\n", output);
}
}
return 0;
}