麻烦编译斐波纳契程序 [英] trouble compiling fibonacci program

问题描述

大家好，


我在90年代早期摆弄BASIC但是把它留在了

那个。现在我正在努力学习C.我试图在我的书中解决一个

练习，但它很难编译。可以

任何人都告诉我错误信息的含义&我应该怎么做？


谢谢。


#include< stdio.h>

长柜台;


无效主要（无效）

{

长纤维;

double fibn_value;

printf（" Enter fibonacci number："）;

scanf（"％ld"，& fibn）;

fibn_value = calc_fib（fibn）;

printf（" \ nThe％ld fibonacci number is％lf \ n"，fibn，

fibn_value ）;

返回;

}


双calc_fib（长纤维）

{

double fibn_value = 1;

double previous1 = 1;

double previous2 = 0;

if（fibn == 1）return（double）0;

else if（fibn == 2）return（double）1;

for（counter = 3; counter< = fibn;计数器++）{

fibn_value = previous1 + previous2;

previous2 = previous1;

previous1 = fibn_value;

}

返回fibn_value;

}


＃gcc fib00.c

fib00.c：在函数''main''：

fib00.c：6：警告：返回类型'' main''不是''int''

fib00.c：顶层：

fib00.c：17：错误：calc_fib的冲突类型

fib00.c：11：错误：先前的隐含声明

''calc_fib''在这里

＃


-

电子邮件：删除z'并反转其余部分。

hello all,

I fiddled with BASIC in the early 90s but left it at
that. Now I am trying to learn C. I tried to solve an
exercise in my book, but it failes to compile. Can
anyone tell me what the error messages mean & what I
should do?

thanks.

#include <stdio.h>
long counter;

void main(void)
{
long fibn;
double fibn_value;
printf("Enter fibonacci number: ");
scanf("%ld", &fibn);
fibn_value = calc_fib(fibn);
printf("\nThe %ld fibonacci number is %lf\n", fibn,
fibn_value);
return;
}

double calc_fib(long fibn)
{
double fibn_value = 1;
double previous1 = 1;
double previous2 = 0;
if(fibn == 1) return (double) 0;
else if(fibn == 2) return (double)1;
for(counter = 3; counter <= fibn; counter++) {
fibn_value = previous1 + previous2;
previous2 = previous1;
previous1 = fibn_value;
}
return fibn_value;
}

# gcc fib00.c
fib00.c: In function ''main'':
fib00.c:6: warning: return type of ''main'' is not ''int''
fib00.c: At top level:
fib00.c:17: error: conflicting types for ''calc_fib''
fib00.c:11: error: previous implicit declaration
of ''calc_fib'' was here
#

--
email: remove z''s and reverse the rest.

推荐答案

Santosh Krisnan写道：

Santosh Krisnan wrote:

你好，


我在90年代早期摆弄BASIC但是留在了

那个。现在我正在努力学习C.我试图在我的书中解决一个

练习，但它很难编译。可以

任何人都告诉我错误信息的含义&我应该怎么做？


＃gcc fib00.c

fib00.c：在函数''main''中：

fib00.c：6：警告：返回类型''main''不是''int''

hello all,

I fiddled with BASIC in the early 90s but left it at
that. Now I am trying to learn C. I tried to solve an
exercise in my book, but it failes to compile. Can
anyone tell me what the error messages mean & what I
should do?

# gcc fib00.c
fib00.c: In function ''main'':
fib00.c:6: warning: return type of ''main'' is not ''int''

int main（ void）

int main(void)

fib00.c：顶级：

fib00.c：17：错误：''calc_fib''的冲突类型

fib00.c：11：错误：先前的隐含声明

''calc_fib''在这里

fib00.c: At top level:
fib00.c:17: error: conflicting types for ''calc_fib''
fib00.c:11: error: previous implicit declaration
of ''calc_fib'' was here

在main之前没有calc_fib的原型，所以编译器

假定函数返回int。要么提供原型，要么交换

main和calc_fib。


-

Ian Collins。

There isn''t a prototype for calc_fib before main, so the compiler
assumes the function returns int. Either provide a prototype, or swap
main and calc_fib.

--
Ian Collins.

Santosh Krisnan写道：

Santosh Krisnan wrote:

hello all，


我在BASIC中摆弄90年代初，但它留在了

那个。现在我正在努力学习C.我试图在我的书中解决一个

练习，但它很难编译。可以

任何人都告诉我错误信息的含义&我应该怎么做？


谢谢。


#include< stdio.h>

长柜台;

hello all,

I fiddled with BASIC in the early 90s but left it at
that. Now I am trying to learn C. I tried to solve an
exercise in my book, but it failes to compile. Can
anyone tell me what the error messages mean & what I
should do?

thanks.

#include <stdio.h>
long counter;

使用静态局部变量而不是全局变量。它减少了程序其他部分的无意中的修改（对于这样一个简单的例子来说不是必需的，但是当你的程序得到它时它会变得很有用>
更大）。

Use a static local variable instead of a global. It reduces inadvertant
modifications by other parts of the program, (not necessary for such a
trivial example, but it''ll become useful when your programs get
bigger).

void main（void）

void main(void)

标准C定义main（ ）作为int main（void）或int main（int

argc，char ** argv）。任何其他签名都不可移植。当然你

可以使用你自己的名字而不是argc和argv，但它们是规范的。

另外** argv可以写成* argv []。

Standard C defines main() as either int main(void) or int main(int
argc, char **argv). Any other signature is not portable. Of course you
can use your own names instead of argc and argv, but they''re canonical.
Also **argv can be written as *argv[].

{

long fibn;

double fibn_value;

printf（" ;输入斐波纳契数：）;

{
long fibn;
double fibn_value;
printf("Enter fibonacci number: ");

除非输出被换行终止，否则不能保证在stdout（通常是屏幕）上立即显示

。或者拨打

fflush（stdout）。

Unless output is terminated by a newline it''s not guaranteed to appear
on stdout, (typically the screen), immediatly. Alternatively call
fflush(stdout).

scanf（"％ld"，& fibn）;

fibn_value = calc_fib（fibn）;

printf（" \ n％ld斐波纳契数是％lf \ n"，fibn，

fibn_value）;

scanf("%ld", &fibn);
fibn_value = calc_fib(fibn);
printf("\nThe %ld fibonacci number is %lf\n", fibn,
fibn_value);

不要将％lf用于双打。这是非标准的。使用％f。

Don''t use %lf for doubles. It''s non-standard. Use %f.

return;

return;

并在这里返回一个int。

And return an int here.

}


double calc_fib（long fibn）

{

double fibn_value = 1;

double previous1 = 1;

double previous2 = 0;

if（fibn == 1）return（double）0;

else if（fibn == 2）return（double）1;

for（counter = 3; counter< = fibn; counter ++）{

fibn_value = previous1 + previous2;

previous2 = previous1;

previous1 = fibn_value;

}

返回fibn_value;

}


#gcc fib00.c

fib00.c：在函数''main''：

fib00.c：6：警告：返回类型''main''是不是''int''

fib00.c：顶级：

fib00.c：17：错误：''calc_fib'的冲突类型'

fib00.c：11：错误：先前的隐含声明

''cal c_fib''在这里

＃

}

double calc_fib(long fibn)
{
double fibn_value = 1;
double previous1 = 1;
double previous2 = 0;
if(fibn == 1) return (double) 0;
else if(fibn == 2) return (double)1;
for(counter = 3; counter <= fibn; counter++) {
fibn_value = previous1 + previous2;
previous2 = previous1;
previous1 = fibn_value;
}
return fibn_value;
}

# gcc fib00.c
fib00.c: In function ''main'':
fib00.c:6: warning: return type of ''main'' is not ''int''
fib00.c: At top level:
fib00.c:17: error: conflicting types for ''calc_fib''
fib00.c:11: error: previous implicit declaration
of ''calc_fib'' was here
#

编译器需要找到一个函数的原型，如果它被称为
在定义之前
。在这里你可以在main（）中调用calc_fib（），但它只是稍后定义的
，因此编译器假定默认返回int，

与它冲突'定义。


将原型放在任何函数定义之前，就在

标题包含之后。

The compiler needs to find the prototype of a function if it''s called
before it''s definition. Here you call calc_fib() in main() but it''s
defined only later, hence the compiler assumes a default return of int,
which conflicts with it''s definition.

Place the prototype before any function definitions, right after the
header includes.

santosh说：

santosh said:

Santosh Krisnan写道：

Santosh Krisnan wrote:

< snip>

<snip>

>>
#include< stdio.h>
long counter;

>>
#include <stdio.h>
long counter;

使用静态局部变量而不是全局变量。

Use a static local variable instead of a global.

再看一遍。它不需要是静态的。


< snip>

Look again. It doesn''t need to be static.

<snip>

>

>

> {
long fibn;
double fibn_value;
printf（" Enter fibonacci number："）;

>{
long fibn;
double fibn_value;
printf("Enter fibonacci number: ");

除非输出被换行终止，否则不能保证在stdout（通常是屏幕）上立即出现

。或者打电话给

fflush（标准输出）。

Unless output is terminated by a newline it''s not guaranteed to appear
on stdout, (typically the screen), immediatly. Alternatively call
fflush(stdout).

正确。

Right.

>

>

> scanf（"％ld"，& fibn）;

> scanf("%ld", &fibn);

到OP：检查返回值。 scanf可能会失败。

To the OP: check the return value. scanf can fail.

> fibn_value = calc_fib（fibn）;
printf（" \\\
％ld fibonacci number is％lf \ n"，fibn，
fibn_value）;

> fibn_value = calc_fib(fibn);
printf("\nThe %ld fibonacci number is %lf\n", fibn,
fibn_value);

不要将％lf用于双打。这是非标准的。使用％f。

Don''t use %lf for doubles. It''s non-standard. Use %f.

为了严格准确，它*现在是标准的，因为C99将其编码。然而，这是b $ b不可携带到C90，这就是世界和他的狗实际上使用的价格。


< snip>


-

Richard Heathfield

Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk

电子邮件：rjh在上述域名中， - www。

To be strictly accurate, it *is* now standard, since C99 codifies it. It is,
however, non-portable to C90, which is what the world and his dog actually
uses.

<snip>

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

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