如何调用仅包含“const char * a”的函数？ [英] How do I call a function taking only "const char *a"
问题描述
大家好，
我对C来说是全新的，并且如何处理不同类型的问题还有一个问题。
的变量。
我有一个你打电话的功能：
double calcdiff（const char * old，const char * new）
>
我启动两个双打，我需要转换才能调用
以上的功能。我该怎么做呢？
谢谢
Hi all,
I''m totally new to C and hava a problem how to handle different types
of variables.
I have a function which which you call by:
double calcdiff(const char *old, const char *new)
I have at start two doubles that I need to convert in order to call the
above function. How would I go about to do that?
Thanks
推荐答案
RoughRyde写道：
RoughRyde wrote:
我对C来说是全新的，并且有一个问题如何处理不同类型
的变量。
I''m totally new to C and hava a problem how to handle different types
of variables.
[获得一本体面的C书。通过新闻组进行滴水是很慢的。]
[Get a decent C book. Dripfeeding via newsgroup is slow.]
我有一个你可以调用的函数：
double calcdiff（const char * old，const char * new）
I have a function which which you call by:
double calcdiff(const char *old, const char *new)
你有一个名为`calcdiff`的函数，它返回一个`double`和
它需要两个/ strings /作为参数？
有些东西坏了。
You have a function called `calcdiff` which returns a `double` and
it takes two /strings/ as arguments?
Something is broken.
我在开始时需要转换两个双打为了调用
以上的功能。我该怎么做呢？
I have at start two doubles that I need to convert in order to call the
above function. How would I go about to do that?
如果你想获得双值的字符串表示，
可能你应该使用`sprintf`。但我先申请伯明翰
原则[1]。如果`calcdiff`需要两个代表
的字符串，将它们变成双精度数，减去它们，然后返回
结果，那么它就是一个很好的naff函数（如果它不是，它是'b $ b给了一个naff名字：你失败的头，我赢的尾巴）。
[1]"我不会从这里开始。
-
Chris" Perikles胜利 Dollin
你服务的对象是谁，你信任谁？ / Crusade /
If you want to get the string representation of a double value,
probably you should use `sprintf`. But I''d apply the Birmingham
principle [1] first. If `calcdiff` takes two strings representing
doubles, turns them into doubles, subtracts them, and returns the
result, then it''s a pretty naff function (and if it doesn''t, it''s
been given a naff name: heads you lose, tails I win).
[1] "I wouldn''t start from here."
--
Chris "Perikles triumphant" Dollin
"Who do you serve, and who do you trust?" /Crusade/
" RoughRyde" < fr ********** @ gmail.comwrote in message
news：11 ******************** *@16g2000cwy.googlegrou ps.com ...
"RoughRyde" <fr**********@gmail.comwrote in message
news:11*********************@16g2000cwy.googlegrou ps.com...
大家好，
我是C的新手和hava一个问题如何处理不同类型
的变量。
我有一个函数，你可以通过以下方式调用：
double calcdiff（const char * old，const char * new）
我启动两个双打，我需要转换才能调用
以上功能。我该怎么做呢？
谢谢
Hi all,
I''m totally new to C and hava a problem how to handle different types
of variables.
I have a function which which you call by:
double calcdiff(const char *old, const char *new)
I have at start two doubles that I need to convert in order to call the
above function. How would I go about to do that?
Thanks
你还没有给出足够的信息。这个功能实际上是什么？
呢？输入应该是什么格式？
例如：
d = calcdiff（" 3.555"，" 1.7E） -16"）;
或
d = calcdiff（三个半，pi）;
如果它只是前者，并且该功能正如其名称
所暗示的那样，为什么不用它：
d = var1 - var2;
-
Fred L. Kleinschmidt
波音助理技术研究员
技术架构师，软件重用项目
You haven''t given enough information. What does the function actually
do? What format are the inputs supposed to be in?
For example:
d = calcdiff( "3.555", "1.7E-16" );
or
d = calcdiff( "three and a half", "pi" );
If it is just the former, and the function does exactly what its name
implies, why don''t you just use:
d = var1 - var2;
--
Fred L. Kleinschmidt
Boeing Associate Technical Fellow
Technical Architect, Software Reuse Project
我想要的是计算两个日期之间的天数。
鉴于我对CI的了解有限，在网上找到了一个这样的例子
，如下所示。问题是我给了indata作为双打。
所以为了轻松解决这个问题，我只想转换那些
双打。
但是如果你对如何解决这个问题有任何其他的建议，那就是 b
赞赏。
double calcdiff（const char * old，const char * new）
{
struct tm t1 = {0};
struct tm t2 = {0};
time_t tt1 = {0};
time_t tt2 = {0};
stotm（& t1，old） ;
stotm（& t2，new）;
tt1 = mktime（& t1）;
tt2 = mktime（& t2 ）;
if（tt1！= -1&& tt2！= -1）
{
long day = 0;
double d = difftime（tt1，tt2）;
if（d <0）d * = -1.0;
day = d / 86400;
返回日;
}
其他
{
返回0 ;
}
}
void stotm（struct tm * p，const char * s）
{
int i;
struct tm blank = {0};
* p =空白;
for（i = 0;我< 4; i ++）
{
p-> tm_year * = 10;
p-> tm_year + = * s ++ - ''0' ';
}
p-> tm_year - = 1900;
for（i = 0; i< 2; i ++）
{
p-> tm_mon * = 10;
p-> tm_mon + = * s ++ - ''0'';
}
--p-> tm_mon;
p-> tm_mday = 10 *（* s ++ - ''0''） ;
p-> tm_mday + = * s ++ - ''0'';
}
Fred Kleinschmidt skrev：
What I want is to calculate the number of days between two dates.
Given my limited knowledge of C I found an example of this on the web
which looked as below. The problem is that I''m given indata as doubles.
So in order to take the easy way out I just wanted to convert those
doubles.
But if you have any other suggestions on how to solve this that''s
appreciated.
double calcdiff(const char *old, const char *new)
{
struct tm t1 = {0};
struct tm t2 = {0};
time_t tt1 = {0};
time_t tt2 = {0};
stotm(&t1, old);
stotm(&t2, new);
tt1 = mktime(&t1);
tt2 = mktime(&t2);
if(tt1 != -1 && tt2 != -1)
{
long day = 0;
double d = difftime(tt1, tt2);
if(d < 0) d *= -1.0;
day = d / 86400;
return day;
}
else
{
return 0;
}
}
void stotm(struct tm *p, const char *s)
{
int i;
struct tm blank = {0};
*p = blank;
for(i = 0; i < 4; i++)
{
p->tm_year *= 10;
p->tm_year += *s++ - ''0'';
}
p->tm_year -= 1900;
for(i = 0; i < 2; i++)
{
p->tm_mon *= 10;
p->tm_mon += *s++ - ''0'';
}
--p->tm_mon;
p->tm_mday = 10 * (*s++ - ''0'');
p->tm_mday += *s++ - ''0'';
}
Fred Kleinschmidt skrev:
" RoughRyde" < fr ********** @ gmail.comwrote in message
news：11 ******************** *@16g2000cwy.googlegrou ps.com ...
"RoughRyde" <fr**********@gmail.comwrote in message
news:11*********************@16g2000cwy.googlegrou ps.com...
大家好，
我是C的新手和hava一个问题如何处理不同类型
的变量。
我有一个函数，你可以通过以下方式调用：
double calcdiff（const char * old，const char * new）
我启动两个双打，我需要转换才能调用
以上功能。我该怎么做呢？
谢谢
Hi all,
I''m totally new to C and hava a problem how to handle different types
of variables.
I have a function which which you call by:
double calcdiff(const char *old, const char *new)
I have at start two doubles that I need to convert in order to call the
above function. How would I go about to do that?
Thanks
你还没有给出足够的信息。这个功能实际上是什么？
呢？输入应该是什么格式？
例如：
d = calcdiff（" 3.555"，" 1.7E） -16"）;
或
d = calcdiff（三个半，pi）;
如果它只是前者，并且该功能正如其名称
所暗示的那样，为什么不用它：
d = var1 - var2;
-
Fred L. Kleinschmidt
波音助理技术研究员
技术架构师，软件重用项目
You haven''t given enough information. What does the function actually
do? What format are the inputs supposed to be in?
For example:
d = calcdiff( "3.555", "1.7E-16" );
or
d = calcdiff( "three and a half", "pi" );
If it is just the former, and the function does exactly what its name
implies, why don''t you just use:
d = var1 - var2;
--
Fred L. Kleinschmidt
Boeing Associate Technical Fellow
Technical Architect, Software Reuse Project
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