Question

案例1：

int i = 10;
int j = i++;

这里，首先将i的值赋给j，然后将i增加1。

案例2：

int i = 10;
int j = ++i;

这里，首先i增加1，然后分配给j。

那么，如果在前缀形式中首先完成增量操作，那么为什么后缀具有比前缀更高的优先级？

Answer 1

这与优先级无关。（此处）在预增量中，分配的值是副作用发生之前的值。对于预增量，它将是副作用后的值。

int i = 10;
int j = i++;

在递增i的值之前是10.在执行这两个语句之后j=10。

int i = 10;
int j = ++i;

这里的值为11.因为incrementmenet先完成然后分配。

Answer 2

manual的{{1}}个operator页面定义为same priority前/后increment。

! ~ ++ -- + - (type) * & sizeof      right to left

首先分配

后增量规则，然后增加

int j = i++;// here first whatever i value is there that will be assigned to j

预增量规则是第一个增量，然后分配

int j = ++i;//++i itself will change i value & then modfied value will assign to j.

例如考虑下面的例子

#include<stdio.h>
int main()
{
        int x = 10;
        int *p = &x;// assume p holds 0x100
        printf("x = %d *p = %d p = %p\n",x,*p,p);
        ++*p++;
        /** in above statement ++ and * having same precedence,then you should check assocaitivity which is R to L
          So start solving from Right to Left, whichever operator came first , solve that one first 
          first *p++ came , again solve p++ first, which is post increment,so address will be same 
         *0x100  means = 10
         now ++10 means = 11
         **/
        printf("x = %d *p = %d p = %p\n",x,*p,p);
}

为什么postfixed具有比prefixed更高的优先级？

2 个答案: