char * to string [英] char* to string

问题描述

愚蠢的问题，但它会告诉我一些因为我很困惑的事情。

如何将char buf [12]的字母存储到字符串中？

每当我这样做时，当buf存在其创建范围时，

字符串数据就会丢失。


这是当前代码：


char buf [12];

int l = strftime（buf，12，"％d /％m /％y"， & birthDate）;

for（int i = 0; i< l; i ++）birthDateStr [i] = buf [i];

cout<< birthDateStr<< l<< endl;


现在，这可行，但是当我退出范围（birthDateStr是一个类

成员）时，birthDateStr再次突然变空！为什么？


如果你有更好的方法（也许有一个临时的char *而不是一个

堆积的数组），请向我解释一下？


谢谢，


-

- 吉普赛男孩

Silly problem, but it''s going to teach me something cause I''m confused..
How do I store the letters of a char buf[12] into a string?
Whenever I do this, when buf exists the scope of its creation, the
string data is lost.

This is the current code :

char buf[12];
int l = strftime(buf,12,"%d/%m/%y",&birthDate);
for(int i=0;i<l;i++) birthDateStr[i] = buf[i];
cout << birthDateStr << l << endl;

Now, this works, but when I exit the scope (birthDateStr is a class
member), birthDateStr is suddenly empty again! Why?

If you have a better way (perhaps with a temporary char * instead of a
heaped array), please explain it to me?

thanks,

--
- gipsy boy

推荐答案

gipsy boy写道：

gipsy boy wrote:

愚蠢的问题，但它会教我一些因为我很困惑......
将char buf [12]的字母存储到字符串中？
每当我这样做时，当buf存在其创建范围时，
字符串数据就会丢失。

char buf [12];
int l = strftime（buf，12，"％d /％m /％y"& birthDate ）;;
for（int i = 0; i< l; i ++）birthDateStr [i] = buf [i];


为什么这么复杂？

我认为borthDateStr的类型是std :: string


cout<< birthDateStr<< l<< endl;

Silly problem, but it''s going to teach me something cause I''m confused..
How do I store the letters of a char buf[12] into a string?
Whenever I do this, when buf exists the scope of its creation, the
string data is lost.

This is the current code :

char buf[12];
int l = strftime(buf,12,"%d/%m/%y",&birthDate);
for(int i=0;i<l;i++) birthDateStr[i] = buf[i];
Why so complicated?
I assume borthDateStr is of type std::string

birthDateStr = buf;
cout << birthDateStr << l << endl;

-

Karl Heinz Buchegger
kb ****** @ gascad.at

gipsy boy写道：

gipsy boy wrote:

愚蠢的问题，但是它会告诉我一些因为我是b $ b困惑的问题..如何将char buf [12]的字母存储到字符串中？


使用带有char const的std :: string ctor *

每当我这样做时，当buf存在其创建范围时，
字符串数据丢失。

这是当前的代码：

char buf [12];
int l = strftime（buf，12，" ％d /％m /％y"& birthDate）;
for（int i = 0; i< l; i ++）birthDateStr [i] = buf [i];
cout<< ; birthDateStr<< l<< endl;

现在，这有效，但是当我退出范围（birthDateStr是一个类
成员）时，birthDateStr再次突然变空！为什么？

Silly problem, but it''s going to teach me something cause I''m confused.. How do I store the letters of a char buf[12] into a string?
Use the std::string ctor which takes a char const*
Whenever I do this, when buf exists the scope of its creation, the
string data is lost.

This is the current code :

char buf[12];
int l = strftime(buf,12,"%d/%m/%y",&birthDate);
for(int i=0;i<l;i++) birthDateStr[i] = buf[i];
cout << birthDateStr << l << endl;

Now, this works, but when I exit the scope (birthDateStr is a class
member), birthDateStr is suddenly empty again! Why?

我怀疑你没有调整birthDateStr的大小，所以没有分配空间

。另一个选择是你正在查看

birthDateStr的副本。我们当然不能肯定。你写当我退出

范围时没有显示范围！


无论如何，birthDateStr = buf出了什么问题; ？

问候，

Michiel Salters

I suspect you have not resized birthDateStr, so there''s no space
allocated. Another option is that you''re looking at a copy of
birthDateStr. We can''t be sure of course. You write "when I exit
the scope" without showing that scope!

Anyway, what was wrong with birthDateStr=buf; ?
Regards,
Michiel Salters

Karl Heinz Buchegger写道：

Karl Heinz Buchegger wrote:

gipsy boy写道：

gipsy boy wrote:

愚蠢的问题，但它会教我一些因为我很困惑的事情......
我该怎么存储这些字母将char buf [12]变成字符串？
每当我这样做时，当buf存在其创建范围时，
字符串数据就会丢失。

这是当前代码：

char buf [12];
int l = strftime（buf，12，"％d /％m /％y"，& birthDate）;
for（int i = 0; i< l; i ++）birthDateStr [i] = buf [i];

Silly problem, but it''s going to teach me something cause I''m confused..
How do I store the letters of a char buf[12] into a string?
Whenever I do this, when buf exists the scope of its creation, the
string data is lost.

This is the current code :

char buf[12];
int l = strftime(buf,12,"%d/%m/%y",&birthDate);
for(int i=0;i<l;i++) birthDateStr[i] = buf[i];

为什么这么复杂？
我认为borthDateStr属于类型std :: string

birthDateStr = buf;

Why so complicated?
I assume borthDateStr is of type std::string

birthDateStr = buf;

我这么复杂，因为这也不起作用。

当我按照你的建议行事时，当我退出
$ b时，birthDateStr就不再是了$ b分配功能的范围。


-

- 吉普赛男孩

I made it so complicated because this doesn''t work either.
When I do what you propose, birthDateStr is nothing anymore when I exit
the scope of the assignment function.

--
- gipsy boy

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