sizeof(array)/ sizeof(int) [英] sizeof(array) / sizeof(int)
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问题描述
我在一个函数中声明了一个数组:
Within a function I have declared an array:
int char_count_array[118] = {0};
稍后,我将此数组传递给函数并计算以下内容:
Later on, I pass this array to a function and calculate the following:
int xx = sizeof(char_count_array);
int xy = sizeof(char_count_array)/sizeof(int);
但是,我得到的结果是: xx = 4 xy = 1
However, the result I get is: xx = 4 xy = 1
我想我会得到: xx = 472(118 * 4)xy = 118(472/4)。
有人会知道我在这里做错了吗?
Would anyone know what I am doing wrong here?
推荐答案
如果您通过将其添加到函数中,最有可能以 int *
而不是 int [118]
的形式出现。因此,您的 sizeof
返回的是指针的大小。当您将C数组传递给函数时,通常也要传递元素数。
If you are passing it to a function, it's most likely going in as int*
instead of int[118]
. So your sizeof
is returning the size of a pointer. When you pass C arrays to functions, it's conventional to also pass the number of elements.
my_func( arr, sizeof(arr)/sizeof(arr[0]) );
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