'char *'和'char（*）[100]'有什么区别？

Question

int main()
{
    char word[100];
    char* lowerCase;

    scanf("%s", word);

    lowerCase = toLowerCase(&word);
    printf("%s", lowerCase);
}

char * toLowerCase(char *str)
{
    int i;

    for(i = 0; str[i] != '\0'; ++i)
    {
        if((str[i] >= 'A') && (str[i] <= 'Z'))
        {
            str[i] = str[i] + 32;
        }
    }

    return str;
}

执行上述代码时收到警告。警告是

try.c: In function 'main':
try.c:16:26: warning: passing argument 1 of 'toLowerCase' from incompatible pointer type [-Wincompatible-pointer-types]
  lowerCase = toLowerCase(&word);
                          ^~~~~
try.c:4:7: note: expected 'char *' but argument is of type 'char (*)[100]'
 char* toLowerCase(char *str);

我不明白为什么会出现此警告？如果我将（单词）传递给函数，则没有警告，但是当我执行以下代码时，输​​出是相同的：

printf("%d", word);
printf("%d", &word);

如果地址相同，那么为什么会出现此警告？

Answer 1

char x[100]

数组x衰减到指针：

x-指向char（char *）的指针

&x-指向100个字符的数组的指针（char (*)[100]）;

&x[0]-指向char（char *）的指针

所有这些指针都引用数组的相同开头，仅类型不同。类型很重要！。

您不应该将&x传递给需要（char *）参数的功能。

为什么类型很重要？：

char x[100];

int main()
{
    printf("Address of x is %p, \t x + 1 - %p\t. The difference in bytes %zu\n", (void *)(x), (void *)(x + 1), (char *)(x + 1) - (char *)(x));
    printf("Address of &x is %p, \t &x + 1 - %p\t. The difference in bytes %zu\n", (void *)(&x), (void *)(&x + 1), (char *)(&x + 1) - (char *)(&x));
    printf("Address of &x[0] is %p, \t &x[0] + 1 - %p\t. The difference in bytes %zu\n", (void *)(&x[0]), (void *)(&x[0] + 1), (char *)(&x[0] + 1) - (char *)(&x[0]));
}

结果：

Address of x is 0x601060,    x + 1 - 0x601061   . The difference in bytes 1
Address of &x is 0x601060,   &x + 1 - 0x6010c4  . The difference in bytes 100
Address of &x[0] is 0x601060,    &x[0] + 1 - 0x601061   . The difference in bytes 1

https://godbolt.org/z/SLJ6xn