连接问题 [英] A concatenation problem
问题描述
以下不工作
int num = 0，
char * string;
string =" qwerty";
strcpy（string，num）;
strcpy（string，" .png"）;
The following is not working
int num = 0,
char *string;
string = "qwerty";
strcpy(string, num);
strcpy(string, ".png");
推荐答案
Caroline< pl *** @ letsdothatagain.com>这样说：
Caroline <pl***@letsdothatagain.com> spoke thus:
以下是行不通的
真的不是一个惊喜...
int num = 0，
char * string;
string =" qwerty" ;;
罚款和花花公子，但是......
strcpy（string，num）;
....这是什么？ strcpy的原型是
char * strcpy（char * dst，const char * src）;
你认为什么时候发生什么？假装num是一个const char
*？它做任何有用的概率都很低。
strcpy（string，" .png"）;
The following is not working
Not really a surprise...
int num = 0,
char *string; string = "qwerty";
Fine and dandy, but...
strcpy(string, num);
....what is this? The prototype for strcpy is
char * strcpy( char * dst, const char * src );
What do you suppose happens when you pretend that num is a const char
*? It certainly has a very low probability of doing anything useful.
strcpy(string, ".png");
同样如此。 ..字符串指向一个字符串文字，这不是
保证可以修改。 strcpy''ing到它肯定是一个糟糕的
计划。
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
And likewise... string points to a string literal, which is not
guaranteed to be modifiable. strcpy''ing to it is certainly a bad
plan.
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
" Caroline" < PL *** @ letsdothatagain.com>在消息中写道
news：da ************************** @ posting.google.c om ...
"Caroline" <pl***@letsdothatagain.com> wrote in message
news:da**************************@posting.google.c om...
以下不起作用
int num = 0，
char * string;
string =" qwerty" ;;
strcpy（string，num）;
strcpy（string，" .png"）;
The following is not working
int num = 0,
char *string;
string = "qwerty";
strcpy(string, num);
strcpy(string, ".png");
1.你的字符串指向常量 QWERTY" ;.你不能修改常量字符串的内容
。
2. strcpy（string，num）不符合函数的原型
strcpy（）。它需要两个指向char的参数作为参数。你可能想要
sprintf（）
1. Your string points to constant "qwerty". You cannot modify the contents
of a constant string.
2. strcpy(string, num) does not conform to the prototype of the function
strcpy(). It takes two pointers to char as arguments. You probably want
sprintf()
Caroline写道：
Caroline wrote:
以下是行不通的
int num = 0，
char * string;
string =" qwerty" ;;
strcpy（string，num）;
strcpy（string， " .png"）;
The following is not working
int num = 0,
char *string;
string = "qwerty";
strcpy(string, num);
strcpy(string, ".png");
也不应该，原因有几个。
在你的代码中，`string ''指向一个字符串文字（qwerty），这是
不可修改。
函数strcpy（）需要一个（const）字符*作为它的第二个参数，
你提供的，'num''，一个int。
即使你的用法是正确的，你也是' d将覆盖
字符串'指向的内容，这显然不是你想要的。
查找函数`sprintf（）''。 *那就是你要找的东西。
HTH，
--ag
BTW - 标识符`string''保留用于实现。
-
-
Artie Gold - - 德克萨斯州奥斯汀
Nor should it be, for several reasons.
In your code, `string'' points to a string literal ("qwerty"), which is
not modifiable.
The function strcpy() expects a (const) char * as its second argument,
where you have provided, `num'', an int.
Even if your use had been correct, you''d be overwriting whatever
`string'' points to, which is evidently not what you want.
Look up the function `sprintf()''. *That''s* what you''re looking for.
HTH,
--ag
BTW - the identifier `string'' is reserved for the implementation.
--
--
Artie Gold -- Austin, Texas
这篇关于连接问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
