C语言 将 char 指针分配给 char 和 char 数组变量
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15424096/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
assigning char pointer to char and char array variable
提问by Majid alDosari
Why is the following ok?
为什么下面没问题？
char *a;
char b[]="asdf";
a=b;
But the following is not?
但是下面是不是？
char a[10];
char b[]="asdf";
a=b;
The above gives error: incompatible types in assignment.
以上给出了错误：赋值中的类型不兼容。
回答by Barath Ravikumar
Both are not ok.
两个都不行。
Maybe you were attempting ,
也许你正在尝试，
char *a;
char b[]="asdf";
a=b;
回答by rashok
char a;
char b[]="asdf";
a=b;
Here you are assigning address of array bto awhich is of chartype. Size of address will be 4 bytes (8 bytes in 64 bit m/c) which you are assigning to 1 byte charvariable aso the values will get truncated. This is legal but its of no use.
在这里，你正在分配数组的地址b到a这是char类型。地址的大小将为 4 字节（64 位 m/c 中的 8 字节），您将其分配给 1 字节char变量，a因此值将被截断。这是合法的，但没有用。
I think you are actually trying to assign first character of barray to a. In that case do a = b[0].
我认为您实际上是在尝试将b数组的第一个字符分配给a. 在那种情况下做a = b[0]。
回答by wats nothin
The value of an array evaluates to the address of the first element within the array. So basically, it's a constant value. That's why when you try to do a=b in the second example, you're trying to do something similar to 2=7, only you have two addresses instead of 2 integers.
数组的值计算为数组中第一个元素的地址。所以基本上，它是一个常数值。这就是为什么当您尝试在第二个示例中执行 a=b 时，您正在尝试执行类似于 2=7 的操作，只有您有两个地址而不是两个整数。
Now it makes sense that the first example would work, since assigning an address to a pointer is a valid operation.
现在第一个示例可以工作是有道理的，因为将地址分配给指针是一个有效的操作。
回答by Marlon Louis
You need to include the below header for string library.
您需要为字符串库包含以下标头。
#include <string.h>
#include <string.h>
Using strcpy(strX, strY);will copy string Y into string X, given there is enough space.
如果strcpy(strX, strY);有足够的空间，使用会将字符串 Y 复制到字符串 X 中。
回答by Kishore
When you say
当你说
char a[10];
'a' is actually
'a' 实际上是
char * const a = malloc(sizeof(char) * 10); // remember to free it, can use alloca() instead
and 'a' is initialized to point to 10 * sizeof(char) of allocated memory.
并且 'a' 被初始化为指向已分配内存的 10 * sizeof(char)。
So
所以
a[1] = 't';
*(a + 1) = 't';
are allowed. But
被允许。但
char *b = "some string";
a = b;
is not allowed.
不被允许。
回答by Digvijay Rathore
Please use strcpy_s function of c++, it is having a syntax of &dest,*source it might help.
请使用 c++ 的 strcpy_s 函数，它的语法为 &dest,*source 可能会有所帮助。
