C++ 错误：将“const char[5]”赋值给“char[10]”时类型不兼容
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Error:incompatible types in assignment of 'const char[5]' to 'char[10]'
提问by Ava
I have defined c as
我已经将 c 定义为
char c[][10]
in function definition and used it like c[i]="gray";
在函数定义中并使用它 c[i]="gray";
Whats wrong? I searched on net, it shows the same syntax.
怎么了？我在网上搜索，它显示了相同的语法。
Thanks.
谢谢。
回答by aschepler
You cannot use assignment (=) on an array. If you change cto an array of pointers, that might work, depending on what you need to do with it.
不能=在数组上使用赋值 ( )。如果您更改c为指针数组，这可能会起作用，具体取决于您需要对它做什么。
const char *c[20];
c[i] = "gray";
Or if the declared type must be array of arrays, you could use strncpy:
或者，如果声明的类型必须是数组数组，则可以使用strncpy：
char c[20][10];
strncpy(c[i], "gray", sizeof(c[i]));
回答by Adam Rosenfield
The problem is that arrays are not assignable in C. String constants like "gray"are character array constants: in this case, the type is char[5](4 + 1 for the terminating null).
问题是数组在 C 中是不可赋值的。字符串常量就像"gray"字符数组常量：在这种情况下，类型是char[5]（4 + 1 表示终止空值）。
If you know that the destination array is large enough to hold the desired string, you can use strcpyto copy the string like so:
如果您知道目标数组足够大以容纳所需的字符串，您可以strcpy像这样复制字符串：
// Make sure you know that c[i] is big enough!
strcpy(c[i], "gray");
A better idea is to use a safer function such as strlcpy(BSD-based systems and Mac OS X) or strcpy_s(Windows):
更好的主意是使用更安全的功能，例如strlcpy（基于 BSD 的系统和 Mac OS X）或strcpy_s（Windows）：
strlcpy(c[i], "gray", 10); // 10 is the size of c[i]
However, these functions are platform-specific and not all that portable. You could also roll your own implementation if speed is not an issue:
但是，这些功能是特定于平台的，并不是那么便携。如果速度不是问题，您也可以推出自己的实现：
size_t strlcpy(char *dst, const char *src, size_t size)
{
size_t len = 0;
while(size > 1 && *src)
{
*dst++ = *src++;
size--;
len++;
}
if(size > 0)
*dst = 0;
return len + strlen(src);
}
Do notuse strncpy, since it could potentially leave you with a non-null-terminated string
千万不能使用strncpy，因为它可能会留下一个非空结尾的字符串
回答by jonsca
Try using strcpy()(found in the cstringheader) instead of just plain assignment.
尝试使用strcpy()（在cstring标题中找到）而不是简单的赋值。
回答by tony gil
this code will work and make the correct assignements in 3 different ways:
此代码将以 3 种不同的方式工作并进行正确的分配：
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
string myString = "hello my friends from Brazil";
char charOut[myString.size()];
strncpy(charOut, myString.c_str(), myString.size());
std::cout << "Char by strncpy string var " << charOut << std::endl;
const char *charOut2;
charOut2 = "sup my homies in L.A.";
std::cout << "Char by const " << charOut2 << std::endl;
string myString2 = "hallo mein bruder in Berlin";
char charOut3[myString2.size()];
strcpy(charOut3, myString2.c_str());
std::cout << "Char by strcpy string var " << charOut3 << std::endl;
}
runs ok on ubuntu servers. did not test on other systems.
在 ubuntu 服务器上运行正常。没有在其他系统上测试。
