错误:从'const char *'到'char *'的无效转换[-fpermissive] [英] error: invalid conversion from 'const char*' to 'char*' [-fpermissive]

问题描述

以下代码中的赋值 C = R.c_str(); 导致G ++抛出以下错误:

The assignment C=R.c_str(); in the code below causes G++ to throw the following error:

错误:从'const char *'到'char *'的无效转换[-fpermissive]"

error: invalid conversion from 'const char*' to 'char*' [-fpermissive]"

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string R = "killme";
    char *C = new char[100];
    C=R.c_str();
    cout<<*C;
}

为什么这是一个错误，我该如何解决?

Why is this an error and how can I solve it?

推荐答案

代码有两个问题.导致编译问题的主要原因是将 c_str()结果(即 const )分配给变量 C ，即不是 const .编译器将此标记为错误，因为否则您可以这样做:

The code has two problems. The main one, which causes a compile issue, is the assignment of c_str() result, which is const, to variable C, which is not const. The compiler tags this as an error, because otherwise you could do this:

C=R.c_str();
C[2] = 'c';

这将写入内存中的只读区域，从而导致未定义的行为.

which would write to a read-only area in memory, causing undefined behavior.

您可以通过两种方式对其进行修复:

You can fix it in two ways:

• 声明 C 一个 const ，即 const char * C = ... 或
• 将内容复制到分配的空间中.

第一种方法很简单-您可以这样做:

The first approach is simple - you do this:

const char *C = R.c_str();

第二种方法是这样的:

char *C = new char[R.size()+1];
std::strcpy(C, R.c_str());

第二个问题是内存泄漏:您的代码将 new 的结果分配给 C ，但从不删除它.如果使用 strcpy 方法，则需要添加

The second problem is a memory leak: your code assigns C a result of new, but never deletes it. If you use strcpy approach, you need to add

delete[] C;

在程序结尾，一旦使用变量 C 完成.

at the end of your program, once you are done using variable C.

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