C using a pointer to a string in a function

Question

a quick question. I'm trying to write a function which recieves a name of a string and prints its second character, But it won't compile (on http://ideone.com), can you address the issue? I don't see the problem as I am sending an address into the function, and asking to access the character in that address.

#include <stdio.h>

int main(void) {
    char line[4] = "abc";
    test(line);
    return 0;
}

void test(int point) {
    point++;
    printf("%c",*point);
    return;
}

The compile error that I am getting is -

Compilation error   time: 0 memory: 10320 signal:0
prog.c: In function ‘main’:
prog.c:5:2: warning: implicit declaration of function ‘test’ [-Wimplicit-function-declaration]
  test(line);
  ^~~~
prog.c: At top level:
prog.c:9:6: warning: conflicting types for ‘test’
 void test(int point) {
      ^~~~
prog.c:5:2: note: previous implicit declaration of ‘test’ was here
  test(line);
  ^~~~
prog.c: In function ‘test’:
prog.c:11:14: error: invalid type argument of unary ‘*’ (have ‘int’)
  printf("%c",*point);
              ^~~~~~

Why do you use int instead of char* ? Also int can not be dereferenced. — BLUEPIXY
– BLUEPIXY, Commented Mar 12, 2017 at 3:41
If you actually read the words in the error message, they're quite specific. You do have to actually read them, though. They're not just there to take up space on the screen; they actually convey information. — Ken White
– Ken White, Commented Mar 12, 2017 at 3:42

dbush · Accepted Answer · 2017-03-12 03:42:56Z

4

Two problems here.

First, you use the function test before it's declared. So the function is implicitly declared as taking an unspecified number of arguments and returning int, i.e. int test(). This conflict with the definition void test(int point) that appears later.

The second issue is that you pass a char array (which decays to a char *) to test, but the function is expecting an int.

Move the definition of test to before main so that is defined before it's used, and change the point parameter from int to char *.

#include <stdio.h>

void test(char *point) {
    point++;
    printf("%c",*point);
    return;
}

int main(void) {
    char line[4] = "abc";
    test(line);
    return 0;
}

answered Mar 12, 2017 at 3:42

dbush

233k27 gold badges261 silver badges334 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

kr1tzy · Accepted Answer · 2017-03-12 03:45:51Z

0

1) It wants a function prototype or for the function to be declared above your main.

2) The parameter point in

void test (int point)

is a pointer, not an int.

answered Mar 12, 2017 at 3:45

kr1tzy

1482 silver badges15 bronze badges

Collectives™ on Stack Overflow

C using a pointer to a string in a function

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related