I would like to concatenate two string with memcpy. But next memcpy is not working. My expected output is "my name is khan".
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *var1 = "my name";
char *var2= "is khan";
char *f_add[20];
memcpy(f_add,var1, strlen(var1)+1);
memcpy(f_add+8,var2, strlen(var2)+1);
printf("%s", f_add);
return 0;
}
char *f_add[20]; defines an array of two pointers to char. You probably want a simple array of 20 char:
char f_add[20];
Then you need to copy to the right place. Copying to f_add+8 starts writing after the null byte that marks the end of the first string, because that string, “my name” has seven non-null characters and one null terminator. So you need to start the copy on the null character:
memcpy(f_add+7, var2, strlen(var2)+1);
You could also use memcpy(f_add + strlen(f_add), var2, strlen(var2)+1);, although that is effectively what strcpy(f_add, var2) does.
Array of pointers is only holding the address (reference) of the string literals. And your code makes no sense at all.
int main(void)
{
char *var1 = "my name";
char *var2= "is khan";
char *f_add[20];
f_add[0] = var1;
f_add[1] = var2;
printf("%s\n%s", f_add[0], f_add[1]);
return 0;
}
or you need to allocate the space for the strings
int main(void)
{
char *var1 = "my name";
char *var2= "is khan";
char *f_add[20];
f_add[0] = malloc(strlen(var1) + 1);
f_add[1] = malloc(strlen(var2) + 1);
memcpy(f_add[0],var1, strlen(var1)+1);
memcpy(f_add[1],var2, strlen(var2)+1);
printf("%s\n%s", f_add[0], f_add[1]);
return 0;
}
char f_add[20], and making it an array of pointers is accidental or a misunderstanding.
mallocforf_addandchar *f_add[20];is not the right one to do concatenationmemcpyand notstrcpy&strcat?strcpy( f_add, var1 ); strcat( f_add, var2 );(and definef_addcorrectly). PS: You problem (one of all) is this+8- you need+7because 8 is beyond the zero, i.e. second string is ignored.