为什么不能，我们一个新的字符串赋值给一个数组，而是一个指针 [英] why cant we assign a new string to an array, but to a pointer

问题描述

我试图重新分配字符串到pre-初始化数组a []，以及所有我能得到一个错误

i was trying to reassign a string to a pre-initialized array a[], and all i could get was an error

main()
{
    char a[] = "Sunstroke";
    char *b = "Coldwave";

    a = "Coldwave";
    b = "Sunstroke";
    printf("\n %s %s",a,b);
}

[错误]：不兼容的类型分配给键入时'的char [10]'从我搜索了这一点，但未能找到任何理由类型'字符*'..我也试图通过重新声明重新分配它像

[Error]: incompatible types when assigning to type 'char[10]' from type 'char *'.. i searched for this but was unable to find any reason.. i also tried to re-assign it by redeclaration like

char a[] = "Sunstroke";

但它没有工作......

but it didnt worked...

，但在一个指针的情况下，有可能在上述程序..

but in case of a pointer it was possible as in above program..

推荐答案

硬codeD字符串，如Coldwave实际上的char [] （CHAR数组）类型 - 但它的未定义行为的修改它们（的 C99 ：6.4.5.6）。需要注意的是下面的，但 B 仍然是一个的char * （字符指针）：

Hard-coded string literals such as "Coldwave" are actually char[] (char array) types -- but it is undefined behavior to modify them (C99:6.4.5.6). Note that below, however, b is still a char* (char pointer):

char *b = "Coldwave";

要其中的char [] 已被分配。没关系。它比这虽然是不同的：

To which a char[] has been assigned. That's okay. It is different than this though:

char a[] = "Coldwave";

这是一个初始化的的的char [] 。您只能初始化一个变量一次，当它被声明，并且初始化，可以在其中填充通过这样分配一个数组或其他复合型（如结构）的唯一情况。你不能做到这一点，但是：

Which is an initialization of a char[]. You can only initialize a variable once, when it is declared, and initialization is the only circumstance in which you can populate an array or other compound type (such as a struct) via assignment like this. You could not do this, however:

char c[] = a;

由于在赋值的右侧使用时，数组变量指向他们重新present数组，这就是功能为什么的char * B = A 工作

Because when used on the right hand side of an assignment, array variables function as pointers to the array they represent, which is why char *b = a works.

所以原因，你不能从上面的变量做到这一点：

So the reason you can't do this with the variables from above:

a = b;
// or
a = "Sunstroke";

是因为会被分配的char * 到的char [] - 白搭;你只能做它周围的其他方法。

Is because that would be assigning a char* to a char[] -- no good; you can only do it the other way around.

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