#include <stdio.h>
#include <string.h>
int main(){
char a[7]= "car";
char b[7]="yyoug";
strcat(a,b[2]);
puts(a);
return 0;
}
This won't compile. It says "passing argument 2 of 'strcat' makes pointer from integer without a cast." I haven't been taught the use of pointers.
b[2] is of char type but strcat expects its both arguments are of char * type.
Use strncat instead. It will append only one byte, i.e. b[2] to the first string if the third argument passed is 1
strncat(a, &b[2], 1);
If you just want to use strncat because you want to learn how it works, then @hacks example is completly perfect. But if you just want to concat on character to a you can also use
a[3] = b[2];
But please keep in mind, either solution only works, if the the destination array, in your case a is large enough.
Without using malloc this might help explain what's going in
#include <assert.h>
#include <stdio.h>
#include <string.h>
int main(){
//positions 0 1 2 3
//"car" == ['c', 'a', 'r', '\0']
char a[7] = "car";
char b[7] = "yyoug";
//strlen("car") == 3;
//"car"[3] == ['\0'];
a[strlen(a)] = b[2];
assert(a[3] == 'o');
a[strlen(a) + 1] = '\0';
assert(a[4] == '\0');
puts(a);
return 0;
}
b[2]gives the values of the 3rd element in the array.&b[2]gives the address.. That is expected fromstrcat()strcat(a,&b[2]);orstrcat(a,b+2);strncat(a,&b[2],1);