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what is the output of following c program if the input is x
#include<stdio.h>
int test (char c[])
{
if (c=="x")
return 1;
else
return 0;
}
int main(int argv, char argc* [])
{
char c[10];
scanf("%s", c);
printf("%d", test(c));
return 0;
}

What I have tried:

#include<stdio.h>
int test (char c[])
{
if (c=="x")
 return 1;
else
return 0;
}
int main(int argv, char argc* [])
{
char c[10];
scanf("%s", c);
printf("%d", test(c));
return 0;
}
Posted
Updated 5-Oct-20 5:47am

Run it, and see what the output is. If the output isn't what is desired, use the debugger to step through the code to see where you went wrong.
   
As you can see with other solutions, == can not compare 2 chars arrays. You should have known if you had followed a good tutorial.
A search on internet would have told you how to compare 2 char arrays in C.

I guess it is also a good time to learn the debugger:
Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

1.11 — Debugging your program (stepping and breakpoints) | Learn C++[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.

-----
Advice: Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.
C++
#include<stdio.h>
int test (char c[])
{
    if (c=="x")
        return 1;
    else
        return 0;
}
int main(int argv, char argc* [])
{
    char c[10];
    scanf("%s", c);
    printf("%d", test(c));
    return 0;
}

Indentation style - Wikipedia[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
   
It will be 0 or 1, as that is what the function test returns.
#include<stdio.h>
#include<string.h>

int test (char c[])
{
if (strcmp(c, "x") == 0)
  return 1;
else
  return 0;
}

int main()
{
  char c[10];
  scanf("%s", c);
  printf("%d", test(c));
  return 0;
}
   
v2
Quote:
int main(int argv, char argc* [])
should be instead
C
int main(int argv, char * argc [])

After fixing the syntax error, I was able to verify that your program always prints 0, even if the user inputs 'x', because in the following line
Quote:
c=="x"
You are comparing apples with oranges.


Try
C
#include <stdio.h>
#include <string.h>

int test (const char * c)
{
  if ( strcmp(c,"x") == 0)
    return 1;
  else
    return 0;
}

int main(int argv, char * argc [])
{
  char c[10];

  int rc= 0;

  if ( scanf("%9s", c ) != 1)
  {
    rc = -1;
    printf("invalid input\n");
  }
  else
    printf("%d\n", test(c));

  return rc;
}
   

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