关于在函数中使用sizeof() [英] About using sizeof() in function

问题描述

你好！
最近，我学习C语言，遇到了问题.示例如下:

Hello!
Recently, I learn C language, I have a problem. The example as follow:

// function prototype
int numOfArray(int num[]);
int main()
{
    int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};

    printf("the number of array is: %d.\n", sizeof(a)/sizeof(a[0]));
    printf("\n");
    printf("after calling function.\n");
    printf("the number of array is: %d.\n", numOfArray(a));

    return 0;
}

int numOfArray(int num[])
{
    return sizeof(num)/sizeof(num[0]);
}

结果是:数组数为:9.
调用函数后.
数组的数量为:1.
众所周知，数组是通过地址在函数中传输的.为什么两个结果不同?我希望有一个人可以帮助我.

The result is: the number of array is:9.
after calling function.
the number of array is:1.
As we know, array is transmitted in function through address. Why the two results are different? I hope someone can help me. Thanks.

推荐答案

sizeof表达式的值是在编译期间确定的，而int numOfArray(int num[])中的一个无法计算出原始大小. num是那里的指针.

The value of a sizeof expression is determined during compiling period, and the one in int numOfArray(int num[]) can not figure out the original size. num is pointer there.

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