为什么 post 通过指针变量 (*variablePointer++) returns 增加一个变量的值是垃圾值？

Question

我只是在玩 pre/post increment/decrement C 语言。在下面的程序中，变量 var 一切正常。但是指针变量自增*varAddress++return垃圾值

#include <stdio.h>
int main(int argc, const char * argv[]) 
{
    int var = 1;
    int *varAddress = &var;
    printf("Value befor pre increment %d\n",*varAddress);
    ++*varAddress;
    printf("Value after pre increment %d\n",*varAddress);
    printf("Value befor post increment %d\n",*varAddress);
    *varAddress++;
    printf("Value after post increment %d\n",*varAddress);
    return 0;
}

输出

Value befor pre increment 1
Value after pre increment 2
Value befor post increment 2
Value after post increment 1606416400
Program ended with exit code: 0

Answer 1

++ 的优先级高于 * 因此通过执行 *varAddress++ 您将指针移动到某个无主位置并尝试取消引用它，这将导致未定义的行为.

Answer 2

根据Operator Precedence，后缀自增的优先级高于间接运算符，所以*varAddress++等价于：

*(varAddress++);

这将增加指针本身，然后指向其他地方未分配的内存，这就是为什么 *varAddress 将 return 垃圾值（这是 UB）。

您可能想要：

(*varAddress)++;

Answer 3

#include<stdio.h>
void main(){
char arr[] ="abcd";
char *p=arr,*q=arr;
char k,temp;
temp = *p++; /* here first it assigns value present in address which
is hold by p and then p points to next address.*/
k = ++*q;/*here increments the value present in address which is 
hold by q and assigns to k and also stores the incremented value in the same 
address location. that why *q will get 'h'.*/
printf("k is %c\n",k); //output: k is h
printf("temp is %c\n",temp);//output: temp is g
printf("*p is %c\n",*p);//output: *p is e
printf("*q is %c",*q);//output: *q is h
}

为什么 post 通过指针变量 (*variablePointer++) returns 增加一个变量的值是垃圾值？

Why post increment to value of a variable by pointer variable (*variablePointer++) returns garbage value?

c

c++

pointers

post-increment

输出