在 C 中调用函数之前的参数评估顺序 [英] Parameter evaluation order before a function calling in C

问题描述

在 C 中调用函数时，是否可以假定函数参数的求值顺序?根据下面的程序，我执行的时候好像没有特定的顺序.

Can it be assumed a evaluation order of the function parameters when calling it in C ? According to the following program, it seems that there is not a particular order when I executed it.

#include <stdio.h>

int main()
{
   int a[] = {1, 2, 3};
   int * pa; 

   pa = &a[0];
   printf("a[0] = %d	a[1] = %d	a[2] = %d
",*(pa), *(pa++),*(++pa));
   /* Result: a[0] = 3  a[1] = 2    a[2] = 2 */

   pa = &a[0];
   printf("a[0] = %d	a[1] = %d	a[2] = %d
",*(pa++),*(pa),*(++pa));
   /* Result: a[0] = 2  a[1] = 2     a[2] = 2 */

   pa = &a[0];
   printf("a[0] = %d	a[1] = %d	a[2] = %d
",*(pa++),*(++pa), *(pa));
   /* a[0] = 2  a[1] = 2 a[2] = 1 */

}

推荐答案

不，函数参数在 C 中没有按照定义的顺序求值.

No, function parameters are not evaluated in a defined order in C.

参见 Martin York 对 C++ 程序员应该知道哪些常见的未定义行为?.

See Martin York's answers to What are all the common undefined behaviour that c++ programmer should know about?.

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