C字符串不从函数返回 [英] C Strings not returning from a function
问题描述
我已经做了一个小程序来演示一个问题我正在用C修复
字符串。
我需要能够从文本行中删除HTML标记。
我创建我的变量，将其传递给我的函数，验证数据已经正确传递了
然后无法获取数据！
如果我将返回值更改为某个随机字符串文字，它会返回
罚款。
所有建议都表示赞赏。
#include< stdio.h>
#include< string.h>
char * FixString（char * strIn）
{
char * strMsg = strIn;
char strTmp [40 ] ="" ;;
char * strReturn =& strTmp;
int i = 0;
int j = strlen（strMsg）;
for（i = 0; i< = j; ++ i）
{
/ *删除usenet上简洁的清理代码* /
strTmp [i] = strMsg [i];
}
printf（ "％s \ n"，strReturn）;
/ *完美打印出字符串* /
返回strReturn; / *更改为返回全部清除它工作正常
* /
}
int main（无效）
{
char * strOut ="这是来自网页的一行< br>";
char * strBack = FixString（strOut）;
printf（"％s \ n"，strBack）; / *打印垃圾 - 为什么？ * /
返回0;
}
I''ve made a small program to demonstrate one problem I''m having fixing
strings in C.
I need to be able to remove HTML mark-ups from text lines.
I create my variable, pass it to my function, verify that the data has
been passed correctly and then cannot get the data back!
If I change the return to some random string literal, it comes back
fine.
All advice appreciated.
#include <stdio.h>
#include <string.h>
char * FixString(char *strIn)
{
char *strMsg = strIn;
char strTmp[40] = "";
char *strReturn = &strTmp;
int i = 0;
int j = strlen(strMsg);
for (i = 0; i <= j; ++i)
{
/* removing the cleanup code for brevity on usenet */
strTmp[i] = strMsg[i];
}
printf("%s\n", strReturn);
/* Prints out the string perfectly */
return strReturn; /* change to return "All clear" and it works fine
*/
}
int main(void)
{
char *strOut = "This is a line from a web page <br>";
char *strBack = FixString(strOut);
printf("%s\n", strBack); /* Prints garbage - why? */
return 0;
}
推荐答案
pkirk25写道：
pkirk25 wrote:
我已经制作了一个小程序来演示一个问题我正在用C修复
字符串。
我需要能够从文本行中删除HTML标记。
我创建变量，将其传递给我的函数，验证数据是否已正确传递
然后无法获取数据！
如果我将返回值更改为某个随机字符串文字，则会返回
罚款。
所有建议都表示赞赏。
#include< stdio.h>
#include< string.h>
char * FixString（char * strIn）
{
char * strMsg = strIn;
char strTmp [40] ="" ;;
char * strReturn =& strTmp;
I''ve made a small program to demonstrate one problem I''m having fixing
strings in C.
I need to be able to remove HTML mark-ups from text lines.
I create my variable, pass it to my function, verify that the data has
been passed correctly and then cannot get the data back!
If I change the return to some random string literal, it comes back
fine.
All advice appreciated.
#include <stdio.h>
#include <string.h>
char * FixString(char *strIn)
{
char *strMsg = strIn;
char strTmp[40] = "";
char *strReturn = &strTmp;
你的编译器没有给你一个有用的诊断这里？如果没有，请将
提高到警告级别或使用更好的编译器。
Didn''t your compiler give you a helpful diagnostic here? If not, turn
up its warning level or use a better compiler.
int i = 0;
int j = strlen（strMsg）;
for（i = 0; i< = j; ++ i）
{
/ *删除清理代码以简化usenet * /
strTmp [i] = strMsg [i];
}
printf（"％s \ n"，strReturn）;
/ *完美打印出字符串* /
返回strReturn ; / *更改为返回全部清除并且工作正常
int i = 0;
int j = strlen(strMsg);
for (i = 0; i <= j; ++i)
{
/* removing the cleanup code for brevity on usenet */
strTmp[i] = strMsg[i];
}
printf("%s\n", strReturn);
/* Prints out the string perfectly */
return strReturn; /* change to return "All clear" and it works fine
你试图返回一个指向本地变量的指针，不要这样做
。传入输出字符串，或者使用动态缓冲区
由函数分配并返回。
-
Ian Collins。
You are attempting to return a pointer to a local variable, don''t do
this. Either pass in the output string, or use a dynamic buffer
allocated by the function and return this.
--
Ian Collins.
[snip]
我的编译器是Visual C ++ 6我知道它已经老了但是它是什么
可用。
我真的认为我的问题是挣扎于间接挣扎
和指针更多的编译器，诱惑它是责备
Microsoft。
[snip]
My complier is Visual C++ 6 wcih I know is old but it is what was
available.
I genuinely think my problems are dwon to struggling with indirection
and pointers moret han the compiler, tempting tho it is to blame
Microsoft.
>
你试图返回一个指向局部变量的指针，不要这样做$ / b $这个。传入输出字符串，或使用函数分配的动态缓冲区
并返回此值。
>
You are attempting to return a pointer to a local variable, don''t do
this. Either pass in the output string, or use a dynamic buffer
allocated by the function and return this.
Ian，如果我重新执行* strReturn指向的字符串数组strTmp，我将获得同样的混乱。
你能用传入输出字符串来解释你的意思吗？什么
我会改变我的代码来做到这一点？很抱歉要求用勺子喂。
Ian, if i retrun the string array strTmp that *strReturn points to, I
get the same mess.
Can you explain what you mean by "pass in the output string"? What
would I change in my code to do that? Sorry to ask to be spoon fed.
pkirk25写道：
pkirk25 wrote:
[snip]
我的编译器是Visual C ++ 6 wcih我知道它已经老了但它是什么
可用。
我真的认为我的问题dwon是为了与间接困难而苦苦挣扎b / b
指针编译器，诱惑它是责怪
微软。
[snip]
My complier is Visual C++ 6 wcih I know is old but it is what was
available.
I genuinely think my problems are dwon to struggling with indirection
and pointers moret han the compiler, tempting tho it is to blame
Microsoft.
C ++编译器应该拒绝编译你剪掉的行;
char * strReturn =& strTmp;
你在这里尝试使用char **初始化char *，使用
char * strReturn = strTmp;
或
char * strReturn =& strTmp [0];
A C++ compiler should refuse to compile the line you snipped;
char *strReturn = &strTmp;
Here you are attempting to initialise a char* with a char**, either use
char *strReturn = strTmp;
or
char *strReturn = &strTmp[0];
>>您正在尝试返回指向局部变量的指针，不要这样做。传入输出字符串，或使用由函数分配的动态缓冲区并返回此信息。
>>You are attempting to return a pointer to a local variable, don''t do
this. Either pass in the output string, or use a dynamic buffer
allocated by the function and return this.
Ian，如果我重新执行* strReturn指向的字符串数组strTmp，我将获得同样的混乱。
你能用传入输出字符串来解释你的意思吗？什么
我会改变我的代码来做到这一点？很抱歉要求用勺子喂食。
Ian, if i retrun the string array strTmp that *strReturn points to, I
get the same mess.
Can you explain what you mean by "pass in the output string"? What
would I change in my code to do that? Sorry to ask to be spoon fed.
添加一个额外的参数：
char * FixString（const char * strIn， char * strOut）
注意将strIn更改为const，你不会在你的
函数中更改它。在你使用strTmp的地方使用strOut。
然后你可以用它来调用它：
const char * strOut ="这是一行来自一个网页< br>"
char * backBuff = char [someSize];
char * strBack = FixString（strOut， backBuff）;
-
Ian Collins。
Add an extra parameter:
char* FixString( const char* strIn, char* strOut )
Note the change of strIn to const, you aren''t changing it in your
function. Use strOut where you use strTmp.
You can then call it with:
const char *strOut = "This is a line from a web page <br>"
char *backBuff = char[someSize];
char *strBack = FixString(strOut, backBuff);
--
Ian Collins.
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