Printing a char variable in C doesnt show properly-CodePudding

So my code is:

#include <stdio.h>

int main() {
    char ch[5] = "funny";
    printf("gum: ");
    printf("ze numbre is %c \n", ch);
}

as far as I learned, it should print:

gum: ze numbre is funny

but instead of outputting the ch variable, if outputs some strange symbol (it looks like a small red square with square FF and sometimes F5 written on it), any tip? I'm coding in VSCode

CodePudding user response：

The problem is the printf conversion specifier %c expects a single character passed as an int such as 'c'. You are passing ch, an array of char, which is automatically converted to a pointer to its first element when passed as a function argument. You should use %s for this argument.

Note also that char ch[5] = "funny"; does not create a proper C string as there is not enough space in ch for the 5 characters in funny and the null terminator. It is safer to let the compiler compute the length of the array with

char ch[] = "funny";

You could also define a char pointer and initialize it to point to a string literal:

char *ch = "funny";

Here is a modified version:

#include <stdio.h>

int main() {
    char ch[] = "array";
    const char *p = "pointer";

    printf("gum: ");
    printf("ch is an %s\n", ch);
    printf("p is a %s\n", p);
    return 0;
}

CodePudding user response：

First I compiled your code :

gcc c.c -Wall
c.c: In function ‘main’:
c.c:7:27: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=]
    7 |     printf("ze numbre is %c \n", ch);
      |                          ~^      ~~
      |                           |      |
      |                           int    char *
      |                          %s

Then changed c to s ( as compiler and Fred Larson said):

   #include<stdio.h>

   int main(){
    char ch[5] = "funny";
    printf("gum: ");
    printf("ze numbre is %s \n", ch);
   }

then recompile ( no errors ) and run

gcc c.c -Wall
./a.out
gum: ze numbre is funny

Works as expected ... but it still not error free as Ed Heal and Fred Larson said: it can give undefined behavior. So next try: let "the compiler to calculate the length of the array"

#include<stdio.h>

int main(){

    char ch[] = "funny";
    printf("gum: ");
    printf("ze numbre is %s \n", ch);
    
}

You can do it in explicit way

char ch[6] = {'f', 'u', 'n', 'n', 'y', '\0'};

or manually udjust length of the array ( include one place for null sign, which is is added automatically at the end of each string in C.

char ch[6] = "funny";

There is also another way

#include<stdio.h>

int main(){

    char ch[5] = "funny";
    printf("gum: ");
    printf("ze numbre is %.5s \n", ch);
    
}

Is it good ?