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;
}
我在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这个警告? 如果我将 (word) 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这个警告?
最佳答案
char x[100]
数组 x
衰减为指针:
x
- 指向字符的指针 (char *
)
&x
- 指向 100 个字符数组的指针 (char (*)[100]
);
&x[0]
- 指向字符的指针 (char *
)
所有这些指针都引用数组的相同开头,只是类型不同。类型很重要!
您不应将 &x
传递给需要 (char *
) 参数的函数。
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
关于c - 'char *' 和 'char (*) [100]' 有什么区别?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60690774/