#include <stdio.h>

  int main() {
    int x = 3;
    while (x > 0, x--) {
      printf("positive");
    }
    printf("%d", & x);
    return 0;
  }

output - positivepositivepositive1472586508
how I got these numbers in output

CodePudding user response：

what you are trying to do is, printing the value of x after some iteration, fixed code is:

   #include <stdio.h>
    int main() {
    int x=3;
     while (x>0)
      {     printf("positive");
            x--;
      }
        printf("%d",x);
    return 0;
}

Note:

1. formatting is important this would help future coders to understand easily.
2. & -- denotes the address of any variable, while you printing the value of x with the ampersand operator, you actually denotes its address.

CodePudding user response：

You got that number because you were printing memory address of that variable.when & used before any variable that represent address of that variable.In case of printf() dont use it just use variable name.

#include <stdio.h>
int main() {
int x=3;
 while (x>0,x--)
  {     
      printf("positive");
    }
printf("%d",x);
return 0;
}

CodePudding user response：

  int main() {
    int x = 3;
    while (x > 0) {
      printf("positive");
      x--;
    }

I little bit edit your code but output is same. up to this point output is "positivepositivepositive" after that you are trying to print memory address of the x variable. that's why output is undefined number. if you use printf statement to print variable value with & sign it always give the memory address of that variable.

printf("%d", & x);

using & sign this part give memory address of the x variable.

printf("%d", x);

if you use this without & sign it give the value of x variable. (x=0)

CodePudding user response：

The code as it is

#include <stdio.h>

  int main() {
    int x = 3;
    while (x > 0, x--) {
      printf("positive");
    }
    printf("%d", & x);
    return 0;
  }

is a bit strange:

the expression x>0, x-- is the same as x--

The comma here is an operator, the sequence operator. An here's what it means:

• the first expression is executed x > 0, resulting in a number, 0 or 1, and this result is discarded. Since the expression has no other effects it means nothing, but with a delay ;) .
• then the second expression x-- is evaluated, and its result is used to evaluate the while condition. So the while will break when x is 0 a.k.a. false in C.
• The expression has a post-decrement operator so it is always executed, leaving x as -1 at the last printf

A sequence e = a,b,c,d; in C means that all expressions are evaluated, but only the result of the last is considered as the result of the expression

Sure, as noted by others, the & is misplaced at the original code.

The code is the same as

#include <stdio.h>
int main(void)
{
    int x = 3;
    while (x--) printf("positive\n");
    printf("%d", x);
    return 0;
}

that shows

positive
positive
positive
-1

more about sequence operations

#include <stdio.h>
int main(void)
{
    int x = 3;
    while (x--) printf("positive\n");
    printf("    [1] %d\n", x);
    x = x>0, x<0, x   2, x--;
    printf("    [2] %d\n", x);
    x = x>0, x<0, 1;
    printf("    [3]  %d\n", x);
    x = 0, x<0, --x;
    printf("    [4]  %d, ", x);

    if ( x = 0, x<0, x-- )
        printf("true\n");
    else
        printf("false\n");
    printf("    [5]  %d\n", x);

    x = 11,2,3,4,5,6,0;
    printf("    [6]  %d\n", x);
    return 0;
}

shows

positive
positive
positive
    [1] -1
    [2] -1
    [3]  0
    [4]  -1, false
    [5]  -1
    [6]  11

and shows a few cases of the comma operator --- not the same as the comma separator as in arguments or declarations