我得到了这个警告:数组函数参数的sizeof将返回“const char *”的大小，而不是“const char[]”[复制]

Possible Duplicate:
Why sizeof(param_array) is the size of pointer?

可能的复制:为什么sizeof(param_array)是指针的大小?

I'm new to C, I got an warning from clang when compiling my code:

我刚到C，在编译我的代码时收到了clang的警告:

#include<stdio.h>

char *strcpy (char destination[],const char source[]);
int main(void) {
    char str1[] = "this is a very long string";
    char str2[] = "this is a short string";
    strcpy(str2, str1);
    puts(str2);
    return 0;
}
char *strcpy (char destination[], const char source[]) {
    int size_of_array = sizeof source / sizeof source[0];
    for (int i = 0; i < size_of_array; i++) {
        destination[i] = source[i];
    }
    return destination;
}

I don't know what does the following warning mean:

我不知道下面的警告是什么意思:

string_copy_withou_pointer.c:12:29: warning: sizeof on array function parameter
      will return size of 'const char *' instead of 'const char []'
      [-Wsizeof-array-argument]
        int size_of_array = sizeof source / sizeof source[0];
                                   ^
string_copy_withou_pointer.c:11:46: note: declared here
char *strcpy (char destination[], const char source[]) {

Any idea?

任何想法?

4 个解决方案

#1

That's because const char source[] in argument position is just syntactic sugar for const char *source. See, e.g., Steven Summit's C notes.

这是因为在参数位置上的const char源只是const char *源的语法糖。看，例如Steven Summit的C notes。

In this particular case, you'll want to call strlen. When not dealing with strings, pass the size of the array as a separate argument.

在这个特殊的情况下，你需要调用strlen。当不处理字符串时，将数组的大小作为单独的参数传递。

#2

This warning is telling you that if you call sizeof(char[]) you won't get the size of the array but the size of a char* pointer.

这个警告告诉您，如果您调用sizeof(char[])，您将无法得到数组的大小，但是您将得到一个char*指针的大小。

This means that your variable size_of_array will be wrong because it won't represent the size of the real array.

这意味着您的变量size_of_array将是错误的，因为它不能代表实际数组的大小。

#3

I think you're looking for this.

我想你是在找这个。

#4

The size of the array doesn't follow when you pass it to the function. Actually it's passed as a pointer which is why the warning message mentions

当您将数组传递给函数时，数组的大小不会跟随。实际上它是作为一个指针传递的，这就是为什么警告信息会被提到。

will return size of 'const char *'

将返回“const char *”的大小

#1