post自增和post自减运算符在c语言运算符优先级规则中有什么位置

Question

根据此 link, post increment and decrement operators have first place. And this link 中的信息，“举个例子：

foo = *p++;

这里 p 作为表达式的副作用递增，但 foo 取 *(p++) 的值而不是 (*p)++，因为一元运算符将右绑定到左 "。

但是按照这些链接中提到的信息，执行后几乎没有任何反应。

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int i = 1;
    int *iptr;
    iptr = &i;
    int num = *iptr++;//1) In this expression the value of 'i' is assigned to num. And an attempt of post incrementing the address stored in *iptr is done as side effect.
    printf("Num value: %d\n",num);

    printf("Pointer: %d\n",*iptr);//2) The address of *iptr is not incremented. If it was then the value of 'i' would not be printed instead it would print the incremented address itself.

    printf("Post increment: %d\n",*iptr++);//3) The address of *iptr will be post incremented (which didn't happen in case of num). But for now the value of 'i' will be printed.

    printf("After increment: %d\n",*iptr);//4) Now the incremented address stored in *iptr will be displayed as there is no value assigned to that address.
    return 0;
}

在上面的实验中，只有在语句终止符之后才能看到post增量的效果。但是，如果对赋值运算符的右操作数进行 post 增量，即使在语句终止符之后也看不到任何效果。 E.G int num = *iptr++; （如上述实验中所述）。那么 post 递增和递减运算符在运算符优先级规则中究竟有什么位置。

Answer 1

您的代码的问题在于它具有未定义的行为：当您将指针指向单个局部变量时，取消引用递增的指针会产生未定义的行为。

解引用递增指针对于指向数组的指针是明确定义的。

int array[] = {1, 2, 3};
int *iptr = &array[0];
int num = *iptr++;

此外，使用 %d 和取消引用运算符打印 iptr 是不正确的：在将 iptr 转换为 [=16= 之后，您需要使用 %p 打印它], 不取消引用:

printf("Pointer: %p\n", (void*)iptr);
// No asterisk here -----------^

现在一切正常（demo）。

post自增和post自减运算符在c语言运算符优先级规则中有什么位置

What place does post increment and post decrement operators have in the rules of operator precedence in c language

c

operators

operator-precedence