c/c++开发分享C中不同类型的指针有什么区别

你是对的。 所有指针都具有相同的大小，并且它们具有相同的数据类型（内存地址）。 所以，你的问题是合法的。

编译器需要知道指针指向的数据类型，以便能够找到如何处理它。 例如，数组基本上是指向（最好）分配的存储区域的指针。 因此， a[1]是*(a+1)的简写。 但是数组的下一个元素从哪里开始呢？ 除非你的指针有一个类型，否则编译器无法知道这一点。 例如，如果你告诉他指向int （4个字节）并且例如a = 0x100 ，他将知道a+1是0x104 ，因为那是下一个元素的地址。

此外，了解指针指向的数据类型对于了解如何取消引用指针（解释数据）至关重要。

这都是关于静态类型检查和指针算术的。 也许最好通过一个具体的例子来说明。

考虑一下：

 #include  int main( int argc, char *argv[] ) { char x[10]; char *p0 = &x[0]; /* ok */ int *p1 = &x[0]; /* <- type checking, warning #1 */ char (*p2)[10] = &x; /* ok */ int (*p3)[10] = &x; /* <- type checking, warning #2 */ (void)printf( "sizeof(char): %ldn", sizeof( char )); (void)printf( "sizeof(int): %ldn", sizeof( int )); (void)printf( "p0: %p, p0+1: %pn", (void*)p0, (void*)( p0+1 )); (void)printf( "p1: %p, p1+1: %pn", (void*)p1, (void*)( p1+1 )); (void)printf( "p2: %p, p2+1: %pn", (void*)p2, (void*)( p2+1 )); (void)printf( "p3: %p, p3+1: %pn", (void*)p3, (void*)( p3+1 )); return 0; } 

重要 ：使用-Wall编译（ gcc -Wall -o test test.c ）

该程序将编译，但你会得到两个关于不兼容的指针类型的警告，这是正确的。

 % gcc -Wall -o test test.c test.c: In function 'main': test.c:9:21: warning: initialization from incompatible pointer type [enabled by default] int *p1 = &x[0]; /* <- type checking, warning #1 */ ^ test.c:11:21: warning: initialization from incompatible pointer type [enabled by default] int (*p3)[10] = &x; /* <- type checking, warning #2 */ ^ 

现在运行程序：

 % ./test sizeof(char): 1 sizeof(int): 4 p0: 0x7fff9f6dc5c0, p0+1: 0x7fff9f6dc5c1 # + 1 char p1: 0x7fff9f6dc5c0, p1+1: 0x7fff9f6dc5c4 # + 1 int (4 bytes here) p2: 0x7fff9f6dc5c0, p2+1: 0x7fff9f6dc5ca # + 10 chars p3: 0x7fff9f6dc5c0, p3+1: 0x7fff9f6dc5e8 # + 10 ints (40 bytes here) 

在这里你可以观察到对指针算术的影响：虽然所有4个指针都被初始化为相同的值，但是相同的操作会产生完全不同的结果。

把它想象成生活空间的不同。 每个住所都有1个地址，但住宅的大小不同。

 /* _________________________________________ /___________Studio Apartments_____________ | _ _ _ _ _ _ _ _ _ _ | |_|0|_|1|_|2|_|3|_|4|_|5|_|6|_|7|_|8|_|9|_| _________________________________________ /____________2 Bed Apartments_____________ | _ _ _ _ _ | |_|0|_____|1|_____|2|_____|3|_____|4|_____| Note: different endianness may look like: _________________________________________ /____________2 Bed Apartments_____________ | _ _ _ _ _ | |_____|0|_____|1|_____|2|_____|3|_____|4|_| */ typedef studio char; //tell the compiler what a "studio" is like typedef apt2br short; //tell it what a 2 bedroom apartment is like /*Now let's build our apartments at the first available address.*/ studio mystudios[10] = /*the letter people live here :)*/ {'A','B','E','C','I','D','O','F','U','G'}; /*We just told our contractor to build somewhere - record locations for later.*/ studio *LetterStudios=&mystudios; /* Let's say we want to print out all of our letter people residents * and no one has built an apartment complex next to them, so the data * following our last letter person is () a '0' */ printf("%sn", (char *)LetterStudios); /* if nothing is built we may get "ABECIDOFUG", but since we did not add our * own '' terminator then we get whatever happens to be next. This is a * buffer overrun ... we may get "ABECIDOFUG" or worse */ 

让我们看一下2字节的情况：

 /* So let's give ourselves some boundaries and loop through it, but this * case has 1 byte elements, so lets look at a 2 byte array */ apt2br myapts[5] = /*people with magic number names live here :)*/ {0xFEFE, 0xB0B, 0xDADE, 0xABE, 0xBABE}; apt2br *MagicApartments=&myapts; /* to get the number of units in an apartment complex we can always divide the * size of the whole complex by the size of a single unit */ for(int i=0;i