printf function 在 C 编程中如何工作？

Question

I wrote this C program.

#include<stdio.h>
int main()
{
  int k=35;
  printf("\n%d %d %d",k==35,k=50,k>40);
  return 0;
}

and got this output.

0 50 0

Now, I am thinking about how the printf function worked and why the output is 0 50 0.

Answer 1

You are invoking undefined behavior by modifying k and using its value in the same statement, before a sequence point.

While any output is legal and valid, it is possible to understand why 0 50 0 happens in this case:

When a C function is called, its parameters can be evaluated in any order, but they are often evaluated in reverse order (right-most-first, left-most-last).

This is not guaranteed, but is frequently true.

So the 4 parameters are evaluated as:

1.) k>40 (false / 0 , because k is 35)
2.) k=50 (assign 50 to k , and evaluate to 50)
3.) k==35 (false / 0 , because k is 50, which is not 35)
4.) "\n%d %d %d" (string literal)

Then printf prints its parameters: 0, 50, 0

Answer 2

There are two problems here:

1. There's no rule to say which order the arguments in a function call (such as when calling printf ) are evaluated. It could be left-to-right, right-to-left, or some other order.
2. In an expression like this, where you both modify k ( k = 50 ) and use its value ( k == 35 , k > 40 ), there's no rule to say whether the old or the new value is used. (And this uncertainty persists regardless of the order of evaluation.)

Because of the second problem, this expression (this printf call) is undefined . It can do anything, even something wildly inexplicable.

If you're calling printf (or any function) on arguments which have "side effects" like this (that is, where an argument is something like k = 50 , which not only provides a value to pass to the function, but also makes some other change as well, in this case updating k 's value), you need to break things up so that the behavior becomes predictable. In the case of printing several values, it's easy -- just use several, separate printf calls:

printf("%d ", k == 35);
printf("%d ", k = 50);
printf("%d\n", k > 40);

Now the behavior is well-defined, and will probably be what you expected.