C中的* ptr + = 1和* ptr ++之间的区别 [英] Difference between *ptr += 1 and *ptr++ in C

问题描述

我刚刚开始研究C，在做一个有关将指针传递给指针作为函数参数的示例时，我发现了一个问题.

I just started to study C, and when doing one example about passing pointer to pointer as a function's parameter, I found a problem.

这是我的示例代码:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int* allocateIntArray(int* ptr, int size){
    if (ptr != NULL){
        for (int i = 0; i < size; i++){
            ptr[i] = i;
        }
    }
    return ptr;
}

void increasePointer(int** ptr){
    if (ptr != NULL){
        *ptr += 1; /* <----------------------------- This is line 16 */
    }
}

int main()
{
    int* p1 = (int*)malloc(sizeof(int)* 10);
    allocateIntArray(p1, 10);

    for (int i = 0; i < 10; i++){
        printf("%d\n", p1[i]);
    }

    increasePointer(&p1);
    printf("%d\n", *p1);
    p1--;
    free(p1);
    fgets(string, sizeof(string), stdin);
    return 0;
}

当我将*ptr+=1修改为*ptr++时，问题发生在第16行.预期的结果应该是整个数组和数字1，但是当我使用*ptr++时，结果是0.

The problem occurs in line 16, when I modify *ptr+=1 to *ptr++. The expected result should be the whole array and number 1 but when I use *ptr++ the result is 0.

+=1和++之间是否有区别?我以为他们都是一样的.

Is there any diffirence between +=1 and ++? I thought that both of them are the same.

推荐答案

差异是由于运算符优先级造成的.

The difference is due to operator precedence.

后递增运算符++的优先级高于取消引用运算符*的优先级.因此*ptr++等效于*(ptr++).换句话说，后置增量会修改指针，而不是其指向的指针.

The post-increment operator ++ has higher precedence than the dereference operator *. So *ptr++ is equivalent to *(ptr++). In other words, the post increment modifies the pointer, not what it points to.

赋值运算符+=的优先级比取消引用运算符*的优先级低，因此*ptr+=1等效于(*ptr)+=1.换句话说，赋值运算符修改指针指向的值，并且不更改指针本身.

The assignment operator += has lower precedence than the dereference operator *, so *ptr+=1 is equivalent to (*ptr)+=1. In other words, the assignment operator modifies the value that the pointer points to, and does not change the pointer itself.

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