C语言 将一个字符连接成一个字符串
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原文地址: http://stackoverflow.com/questions/16653458/
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Concatenating a char into a string
提问by Bestknighter
I am trying to read a string from the console. But I want to read it char by char. And I'm having trouble with concatenating the char into the string AND with breaking the loop. Here is the code:
我正在尝试从控制台读取字符串。但我想逐个字符地阅读它。而且我在将字符连接到字符串和打破循环时遇到了麻烦。这是代码：
char* createString(){
char c;
char *string;
int x=0;
string = (char*) calloc(1, sizeof(char));
do{
c = getche();
if(c != '\n'){
x++;
realloc(string, sizeof(char)*x);
strcat(string, &c);
};
}while(c != '\n');
return string;
};
When I run this code, each concatenation adds 3 chars, instead of just 1. It's like is accessing non-allocated memory... (For example, if I press a, the final string is a%T. Then, if I press another key, sfor example, string becomes a%Ts%T)
当我运行此代码时，每个连接添加 3 个字符，而不仅仅是 1 个。这就像访问未分配的内存......（例如，如果我按下a，最终字符串是a%T。然后，如果我按下另一个键，s对于例如，字符串变为a%Ts%T)
And when I press Enter, it goes into the if and doesn't get out of the loop.
当我按下 时Enter，它会进入 if 并且不会退出循环。
I have no clue of why and what is going on...
我不知道为什么以及发生了什么......
EDIT
编辑
Based on other tries and responses until now, I changed my code and now it's like this:
根据到目前为止的其他尝试和响应，我更改了代码，现在是这样的：
char* digitarString(){
char c[2];
char *string;
string = (char*) calloc(1, sizeof(char));
do{
c[0] = getche();
c[1] = 'c[0] = getche();
';
if(c[0] != '\n'){
strcat(string, c);
};
}while(c[0] != '\n');
return string;
};
BUT, there are still two issues...
但是，还有两个问题......
- The code works, but I think that it's writing in non-allocated memory.
- When I press
Enterit still doesn't work. It keep entering the loop and if.
- 该代码有效，但我认为它是在未分配的内存中写入的。
- 当我按下
Enter它仍然不起作用。它不断进入循环和如果。
Forget about the Enter... I changed it...
忘记Enter......我改变了它......
scanf("%c", &c[0]);
to
到
strcat(string, &c);
and worked awesomely well.
并且工作得非常好。
回答by u__
Ok here is the solution
好的，这是解决方案
strncat(string, &c,1);
change this to
将此更改为
c = getche();
now the answer to the question why ?
现在这个问题的答案为什么？
well first of call the below statement
首先调用下面的语句
c
+-----------+------------+------------+-----------+------------+------------+
| a | | | | | |
+---------- +---------- +---------- +---------- +--------- - +---------- +
x = &c x+1 x+2 ......
will scan a value for us and will place in variable called c
将为我们扫描一个值并将其放置在名为 c 的变量中
now lets consider the variable is placed in an arbitrary memory location x
现在让我们考虑将变量放置在任意内存位置 x
strcat(string, &c);
now to the next important statement
现在进入下一个重要声明
int x=1;
the second argument above is supposed to be a string means a NULL termination at the end but we can not guarantee that x+1 location is NULL and if x+1 is not NULL then the string will be more than a single character long and it will end up appending all those characters to your original string hence garbage.
上面的第二个参数应该是一个字符串，意味着最后一个 NULL 终止，但我们不能保证 x+1 位置为 NULL，如果 x+1 不是 NULL，那么字符串将超过一个字符长，它最终会将所有这些字符附加到您的原始字符串中，因此是垃圾。
I hope it's clear now...
我希望现在清楚了...
P.S - if you have access to gdb you can check practically..
PS - 如果您可以访问 gdb，您可以实际检查..
回答by MOHAMED
1) you should initialize
1）你应该初始化
realloc(string, sizeof(char)*x);
2) you should update this line:
2）你应该更新这一行：
string = realloc(string, sizeof(char)*x);
to
到
strcat(string, &c);
3) Yoou do not need for strcat to concatenate. So instead of using
3) 你不需要 strcat 来连接。所以而不是使用
string[x-2] = c;
string[x-1] = 'char* createString(void){
int c;
char *string=NULL;
int x=0;
while(1){
c = getche();
string = realloc(string, sizeof(char)*(x+1));
if('\n' != c)//input <ctrl+j> or use getchar()
string[x++] = (char)c;
else {
string[x] = '##代码##';
break;
}
}
return string;
}
';
use the following lines
使用以下几行
##代码##