C++ 在 MessageBox 中显示 int 变量
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Display an int variable in a MessageBox
提问by user3076025
I am working on an old app written in Visual C++ 6.0. I am trying to display an intvariable in a MessageBoxfor debugging reasons. Here is my code, I thought this would be a simple process, but I am just learning C++. The two lines that are commented I have tried as well with similar errors. Below is the error I am getting.
我正在开发一个用 Visual C++ 6.0 编写的旧应用程序。出于调试原因，我试图int在 a 中显示一个变量MessageBox。这是我的代码，我认为这将是一个简单的过程，但我只是在学习 C++。评论的两行我也尝试过类似的错误。下面是我得到的错误。
int index1 = 1;
char test1 = index1;
// char var1[] = index1;
// char *varGo1 = index1;
MessageBox(NULL, test1, "testx", MB_OK);
error C2664: 'MessageBoxA' : cannot convert parameter 2 from 'char' to 'const char *'
错误 C2664：“MessageBoxA”：无法将参数 2 从“char”转换为“const char *”
回答by Ivan Aksamentov - Drop
Why bother with C-style strings if you tagged C++?
如果您标记了 C++，为什么还要打扰 C 风格的字符串？
Although Mark Ransomprovided MFC solution(which is perfectly valid), here is a Standard C++ one:
尽管Mark Ransom提供了MFC 解决方案（这是完全有效的），但这里是标准 C++解决方案：
int index1 = 1;
std::string test1 = std::to_string(index1);
MessageBoxA(NULL, test1.c_str(), "testx", MB_OK);
References:
参考：
Use boost::formatfor more sophisticated formatting.
使用boost::format更复杂的格式。
回答by Mark Ransom
CString str1;
str1.Format(_T("%d"), index1);
MessageBox(NULL, str1, "testx", MB_OK);
CString's Formatworks just like printfto populate the string with the parameter list.
CString 的Format工作就像printf用参数列表填充字符串一样。
回答by Oleh
int index1 = 1;
char buf[10];
itoa(index1,buf,10);
MessageBox(NULL,buf,"Caption",MB_OK);
Can try this
可以试试这个
回答by Bathsheba
The second parameter of MessageBoxneeds to be a pointer to a stringof chars, terminated with NULL. Passing a charwill not work.
的第二个参数MessageBox需要是一个指向s字符串的指针char，以 NULL 结尾。通过 achar将不起作用。
But, learning to use a debugger is an integral part to learning a language. Why not build a debug build and set a breakpoint on char test1 = index1;instead? You do that by pressing F9 when the cursor is on that line.
但是，学习使用调试器是学习语言不可或缺的一部分。为什么不构建调试版本并设置断点char test1 = index1;呢？当光标位于该行上时，您可以通过按 F9 来完成此操作。
回答by Jerry Coffin
For what it's worth, I prefer to use a manipulator for this:
对于它的价值，我更喜欢为此使用操纵器：
#include <sstream>
#include <iostream>
#include <windows.h>
using std::ostringstream;
using std::ostream;
ostream &msg_box(ostream &s) {
ostringstream &os = dynamic_cast<ostringstream &>(s);
MessageBox(NULL, os.str().c_str(), "testx", MB_OK);
return s;
}
int main() {
ostringstream msg;
msg << "The number is: " << 10 << msg_box;
return 0;
}
This maintains (mostly) the same interface nearly everybody's already accustomed to with iostreams, avoids the type-unsafe CString::Format, and avoids having several lines of distraction everywhere you're going to display a little information for debugging. The other obvious good point is that if you've overloaded operator<<for your own type, that overload will work with this as well.
这保持了（大部分）相同的界面，几乎每个人都已经习惯了iostreams，避免了类型不安全CString::Format，并避免了在你要显示一些调试信息的任何地方都有几行分心。另一个明显的好处是，如果您operator<<为自己的类型进行了重载，那么该重载也将适用于此。
回答by HaSeeB MiR
Here is the pure Csolution using sprintfmethod to store all input in buffer and passing that buffer to MessageBox.
这是使用sprintf方法将所有输入存储在缓冲区中并将该缓冲区传递给MessageBox的纯C解决方案。
#include <stdio.h>
#include <windows.h>
int main(void)
{
int intVal = 50;
float fltVal = 5.5;
char *str = "Test String";
char buf[1024] = {'const char * test1= new char();
'};//buffer to store formatted input.
//convert formatted input into buffer.
sprintf(buf,"Int value : %d\nFloat value : %f\nString : %s\n",intVal,fltVal,str);
//display whole buffer.
MessageBox(NULL,buf,"INFO",MB_ICONINFORMATION);
return 0;
}
回答by Hymanzawa
Acording to your error, you should declare a const pointer on the second parameter. Like this,
根据您的错误，您应该在第二个参数上声明一个 const 指针。像这样，
std::string test1= "";
MessageBox(NULL, test1.c_str(), "testx", MB_OK);
or use
或使用
##代码##Also using just "Text" will work.
也可以只使用“文本”。
