为什么将 C-Array 传递给函数时 sizeof() 值错误? [英] Why does a C-Array have a wrong sizeof() value when it's passed to a function?

问题描述

完整示例:

#include <stdio.h>

void test(int arr[]) {
    int arrSize = (int)(sizeof(arr) / sizeof(arr[0]));
    printf("%d
", arrSize); // 2 (wrong?!)
}

int main (int argc, const char * argv[]) {
    int point[3] = {50, 30, 12};

    int arrSize = (int)(sizeof(point) / sizeof(point[0]));
    printf("%d
", arrSize); // 3 (correct :-) )

    test(point);

    return 0;
}

在将它传递给函数之前， sizeof 给了我正确的值.对函数中完全相同的数组执行完全相同的操作会产生奇怪的结果.少了一个元素.为什么?

Before passing it to a function, sizeof gives me the correct value. Doing the exact same thing on the exact same array in the function gives weird results. There's one element missing. Why?

推荐答案

当您将数组传递给 C 中的函数时，数组衰减为指向其第一个元素的指针.当您在参数上使用 sizeof 时，您获取的是指针的大小，而不是数组本身.

When you pass an array into a function in C, the array decays into a pointer to its first element. When you use sizeof on the parameter, you are taking the size of the pointer, not the array itself.

如果你需要函数知道数组的大小，你应该把它作为一个单独的参数传递:

If you need the function to know the size of the array, you should pass it as a separate parameter:

void test(int arr[], size_t elems) {
   /* ... */
}

int main(int argc, const char * argv[]) {
   int point[3] = {50, 30, 12};
   /* ... */
   test(point, sizeof(point)/sizeof(point[0]));
   /* ... */
}

还要注意，出于类似的原因(将 sizeof 设为指针)，sizeof(point)/sizeof(point[0]) 技巧不会适用于动态分配的数组，仅在堆栈上分配的数组.

Also note that, for a similar reason (taking the sizeof a pointer), the sizeof(point)/sizeof(point[0]) trick doesn't work for a dynamically allocated array, only an array allocated on the stack.

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