C中的函数调用之前参数评测秩序 [英] Parameter evaluation order before a function calling in C

问题描述

是否可以在C调用时，它承担的函数参数的计算顺序？根据以下程序，似乎没有特定的顺序，当我执行它。

 的#include＆LT;＆stdio.h中GT;诠释的main（）
{
   诠释一个[] = {1，2，3};
   INT * PA;   PA =＆放大器;一个[0];
   的printf（A [0] =％d个\\ TA [1] =％d个\\ TA [2] =％d个\\ N，*（PA），*（PA ++），*（++ PA））;
   / *结果：a [0] = 3 [1] = 2 [2] = 2 * /   PA =＆放大器;一个[0];
   的printf（A [0] =％d个\\ TA [1] =％d个\\ TA [2] =％d个\\ N，*（PA ++），*（PA），*（++ PA））;
   / *结果：a [0] = 2 [1] = 2 [2] = 2 * /   PA =＆放大器;一个[0];
   的printf（A [0] =％d个\\ TA [1] =％d个\\ TA [2] =％d个\\ N，*（PA ++），*（++ PA），*（PA））;
   / *一个[0] = 2 [1] = 2 [2] = 1 * /}
 

解决方案

没有，函数参数没有在C定义的顺序进行。

请参阅马丁纽约的答案<一个href=\"http://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviour-that-c-programmer-should-know-about\">What都是常见的未定义行为的C ++程序员应该了解？。

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

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

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

}

解决方案

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

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

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