C++ & DirextX9 问题~ 帮帮忙下

冰室

#   include   <iostream>   
  #   include   <string>
int num1 = 1;
int num2 = 2;
string   temp;
temp=sprintf(str,"%d",num1)+sprintf(str,"%d",num1);
试试吧

ahwai89

想请教下~
例子:-
int num1 = 1;
int num2 = 2;
char operator = '+';
cout << num1 << a << num2 << " = " << num1+num2  << endl; //console output
在C++要怎样改才能把cout的整句换去string?
我要用dirext的DrawTextA来output..

老师要我们做小游戏~console的就会
当要convert 去 GUI 就头大~~   
帮帮忙下~谢了~

geekman

1. int sprintf ( char * str, const char * format, ... );
   
2. ....
   
3. Return Value
   
4. On success, the total number of characters written is returned. This count does not include the additional null-character automatically appended at the end of the string.
   
5. On failure, a negative number is returned.
   
6. 
   
7. 来自：http://www.cplusplus.com/reference/clibrary/cstdio/sprintf.html

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我相信楼主想要问的是如何把 “[num1] [operator] [num2] = [sum of num1 and num2]" 这个句子化成 C String （或者char *）以便提供给 DrawTextA（）使用。

看了上面的sprintf()的defination，就可发现，冰室大大的答案完全错误。temp的内容将会相当于2，因为每一个sprintf()都会return 1（因为每一个sprintf()都会返回所写出的字元的数量，在这里，num1 和 num2 都只是单个数字），所以 temp = 1 + 1 => temp = 2。

正确答案是：

1. char buffer[MAX_BUFFER]; //MAX_BUFFER can be any number,
   
2. int num1 = 1, num2 = 2;
   
3. char operator = '+';
   
4. 
   
5. sprintf(buffer, "%d %c %d = %d", num1, operator, num2, num1+num2);

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[ 本帖最后由 geekman 于 28-11-2008 11:51 PM 编辑 ]

onlylonly

可以用stringstream

#include <sstream>

.......

stringstream ss;
ss << num1 << a << num2 << " = " << num1+num2  << endl

要 convert string 就
ss.str();
convert c string 就

ss.str().c_str();