I'm Trying to do some simple c programming that will return the char value. The program is running with no error but, the output of the expected string does not appear. What I expected it will return the Halloworld when i run it.
#include <stdio.h>
#include <string.h>
char screen(char c[]);
int main(){
char b[] = "Hallo";
//screen(b);
printf("%s",screen(b));
}
char screen(char c[]){
strcat(c, "world");
return c;
}
The declaration of strcat() returns a pointer to char (char *). So your function screen() must also.
Another thing is, you can't do this char b[] = "Hallo";, because then the character array b is only large enough to handle Hallo. You'll need it larger than Hallo, and the rest will be filled with 0x00, or NUL. Like so:
48 61 6C 6C 6F 20 00 00 00 00 00 00 00 00 00 00 00 00 00 00
And then after you strcat() the characters world onto the end:
48 61 6C 6C 6F 20 77 6F 72 6C 64 00 00 00 00 00 00 00 00 00
The first left over 0x00 will act as a null terminator when passed to printf().
#include <stdio.h>
#include <string.h>
char* screen(char c[])
{
strcat(c, "world");
return c;
}
int main()
{
char b[20] = "Hallo ";
screen(b);
printf("%s",b);
return 0;
}
int is just an int but a string (array of char) is more than just a char. The convention in C is to use a pointer to the first char of the string to represent a string.
char[]tocharand I would expect a compile-time error from that.bis a constant object without enough space for any more data to be appended to it, which means this program is undefined behavior and anything could happen, including nothing at all.-Wall -Werroror whatever the equivalent in your compiler would be: don't just ignore a compiler warning like that.) If you don't understand that warning, then that's a good question in itself (but I'm pretty sure it's been asked and answered thoroughly already here somewhere).