让我们来看看:

#include <stdio.h>
#include <string.h>

int main()
{
    char *p = "hello";
    char q[] = "hello"; // no need to count this

    printf("%zu\n", sizeof(p)); // => size of pointer to char -- 4 on x86, 8 on x86-64
    printf("%zu\n", sizeof(q)); // => size of char array in memory -- 6 on both

    // size_t strlen(const char *s) and we don't get any warnings here:
    printf("%zu\n", strlen(p)); // => 5
    printf("%zu\n", strlen(q)); // => 5

    return 0;
}

Foo *和Foo[]是不同的类型，编译器对它们的处理也不同(pointer =地址+指针类型的表示，array =指针+数组的可选长度，如果已知，例如，如果数组是静态分配的)，详细信息可以在标准中找到。在运行时级别上，它们之间没有区别(在汇编程序中，好吧，几乎没有区别，请参阅下面)。

此外，在C常见问题解答中有一个相关的问题:

Q: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways: As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size). Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching). See also questions 1.31, 6.1, 6.2, 6.8, and 11.8b. References: K&R2 Sec. 5.5 p. 104 ISO Sec. 6.1.4, Sec. 6.5.7 Rationale Sec. 3.1.4 H&S Sec. 2.7.4 pp. 31-2

对于这样的情况，效果是相同的:您最终传递字符串中第一个字符的地址。

声明显然是不一样的。

下面的代码为字符串和字符指针预留内存，然后将指针初始化为指向字符串中的第一个字符。

char *p = "hello";

而下面的方法只为字符串预留内存。所以它实际上可以使用更少的内存。

char p[10] = "hello";

Char p[3] = "hello" ?应该是char p[6] = "hello"记住在C语言的"string"结尾有一个'\0'字符。

不管怎样，数组在C语言中只是指向内存中调整对象的第一个对象的指针。唯一不同的是语义。虽然可以将指针的值更改为指向内存中的不同位置，但创建后的数组将始终指向相同的位置。 此外，当使用数组时，“新建”和“删除”会自动为你完成。

你不允许改变字符串常量的内容，也就是第一个p所指向的内容。第二个p是一个用字符串常量初始化的数组，您可以更改其内容。

Char *和Char[]是不同的类型，但并不是在所有情况下都能立即显示出来。这是因为数组衰减为指针，这意味着如果在需要char*类型的表达式的地方提供了char[]类型的表达式，编译器会自动将数组转换为指向其第一个元素的指针。

你的例子函数printSomething需要一个指针，所以如果你试图像这样传递一个数组给它:

char s[10] = "hello";
printSomething(s);

编译器会假装是你写的:

char s[10] = "hello";
printSomething(&s[0]);

让我们来看看:

#include <stdio.h>
#include <string.h>

int main()
{
    char *p = "hello";
    char q[] = "hello"; // no need to count this

    printf("%zu\n", sizeof(p)); // => size of pointer to char -- 4 on x86, 8 on x86-64
    printf("%zu\n", sizeof(q)); // => size of char array in memory -- 6 on both

    // size_t strlen(const char *s) and we don't get any warnings here:
    printf("%zu\n", strlen(p)); // => 5
    printf("%zu\n", strlen(q)); // => 5

    return 0;
}

Foo *和Foo[]是不同的类型，编译器对它们的处理也不同(pointer =地址+指针类型的表示，array =指针+数组的可选长度，如果已知，例如，如果数组是静态分配的)，详细信息可以在标准中找到。在运行时级别上，它们之间没有区别(在汇编程序中，好吧，几乎没有区别，请参阅下面)。

此外，在C常见问题解答中有一个相关的问题:

Q: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways: As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size). Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching). See also questions 1.31, 6.1, 6.2, 6.8, and 11.8b. References: K&R2 Sec. 5.5 p. 104 ISO Sec. 6.1.4, Sec. 6.5.7 Rationale Sec. 3.1.4 H&S Sec. 2.7.4 pp. 31-2