遇到调用函数的问题 [英] having a problem with calling a function
问题描述
我正在尝试自学编程，我正在通过Ira Pohl的解剖书C进行练习。
我被困在其中一个上并且无法弄清楚如何修复它。
练习是为你编写一个程序，询问用户输入的数字，然后打印出许多素数。提供了确定数字是否为素数的函数。
我的问题是我无法弄清楚如何重复调用函数并只打印从中返回true的数字素数函数。到目前为止我的是这个。
#include" primes.h"
main（无效）
{
int x，y，n;
printf（PRIME NUMBERS将打印出来）;
printf（你想看多少？）;
scanf（"％d"& x）;
is_prime（ n）;
{
for（y = 0; y< = x; ++ y）
printf（"％5d： ％5d \ n"，y，n）
}
}
这里是is_prime.c
#include" primes.h"
int is_prime（int n）
{
int k，limit;
if（n == 2）
返回1;
if（n％2 = = 0）
返回0;
limit = n / 2;
for（k = 3; k <= limit; k + = 2）
if（n％k == 0）
返回0;
返回1;
}
这里是primes.h
#include< stdio.h>
#include< stdlib.h>
int is_prime（int n）;
请帮帮我....这件事真的让我烦恼。
Im trying to teach myself programming, and I am doing exercises out of the book C by Dissection by Ira Pohl.
I am stuck on one of them and can''t figure out how to fix it.
The exercise is for you to write a program that asks for a user inputted number and then print out that many prime numbers. The function that determines if a number is prime or not is provided.
My problem is that I can''t figure out how to call the function repetitively and only print the numbers that return a true from the prime number function. What I have so far is this.
#include "primes.h"
main(void)
{
int x,y,n;
printf("PRIME NUMBERS WILL BE PRINTED\n");
printf("How many do you want to see? ");
scanf("%d",&x);
is_prime(n);
{
for(y=0;y<=x;++y)
printf("%5d: %5d\n",y,n)
}
}
here is is_prime.c
#include "primes.h"
int is_prime(int n)
{
int k, limit;
if (n == 2)
return 1;
if (n % 2 == 0)
return 0;
limit = n / 2;
for (k = 3; k <= limit; k += 2)
if (n % k == 0)
return 0;
return 1;
}
here is primes.h
#include <stdio.h>
#include <stdlib.h>
int is_prime(int n);
Please help me.... this thing is really bugging the crap out of me.
推荐答案
interpim写道：
interpim wrote:
我正在尝试自学编程，而我正在练习本书的练习由Ira Pohl解剖。我被困在其中一个上面并且无法弄清楚如何解决它。
这个练习是为你编写一个程序，要求用户输入
编号，然后打印出许多素数。提供了
确定数字是否为素数的函数。
该函数返回一个值。用它。 （见下文。）
我的问题是我无法弄清楚如何重复调用函数
并且只打印从素数中返回true的数字
数字功能。到目前为止我的是这个。
#include" primes.h"
你忘了：
#include< stdio.h>
这是必需的，因为你正在使用printf。
main（无效）
{x / y，n; n;
printf（" PRIME NUMBERS WILL打印出来的打印（你想看多少？）;
scanf（"％d"& x）;
is_prime （n）;
{
（y = 0; y< = x; ++ y）
printf（"％5d：％5d \ n"，y，n）
}
如果输入值为1，则is_prime函数无法正常工作，因此
将其更改为：
for（y = 2; y< = x; ++ y）
{
if（is_prime（n））
{
printf（"％d是prime \ n"，y）;
}
else
{
printf（"％d是composite.\ n，y）;
}
}
现在，main（）返回int，所以从main（）返回一个int：
返回0;
}
Im trying to teach myself programming, and I am doing exercises out of the
book C by Dissection by Ira Pohl. I am stuck on one of them and can''t
figure out how to fix it.
The exercise is for you to write a program that asks for a user inputted
number and then print out that many prime numbers. The function that
determines if a number is prime or not is provided.
That function returns a value. Use it. (See below.)
My problem is that I can''t figure out how to call the function
repetitively and only print the numbers that return a true from the prime
number function. What I have so far is this.
#include "primes.h"
You forgot:
#include <stdio.h>
This is required, because you are using printf.
main(void)
{
int x,y,n;
printf("PRIME NUMBERS WILL BE PRINTED\n");
printf("How many do you want to see? ");
scanf("%d",&x);
is_prime(n);
{
for(y=0;y<=x;++y)
printf("%5d: %5d\n",y,n)
}
Your is_prime function doesn''t work correctly for an input value of 1, so
change this to:
for(y = 2; y <= x; ++y)
{
if(is_prime(n))
{
printf("%d is prime\n", y);
}
else
{
printf("%d is composite.\n", y);
}
}
Now, main() returns int, so return an int from main():
return 0;
}
-
Richard Heathfield： bi **** @ eton.powernet.co.uk
Usenet是一个奇怪的地方。 - Dennis M Ritchie，1999年7月29日。
C FAQ： http://www.eskimo.com/~scs/C-faq/top.html
K& R答案，C书等：< a rel =nofollowhref =http://users.powernet.co.uk/etontarget =_ blank> http://users.powernet.co.uk/eton
--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
这有意义吗？
main（无效）
{
int x，y，n;
printf（PRIME NUMBERS将被打印）;
printf（"你想要多少个）看？"）;
scanf（"％d"& x）;
for（y = 0; y< = x; ++ y ）
{
n = is_prime（y）;
printf（"％5d：％5d \ n"，y，n） ;
}
}
David
星期二，2003年12月16日04:34： 20 UTC，interpim <乐*** @ atrc.navy.mil>写道：
Does this make sense?
main(void)
{
int x,y,n;
printf("PRIME NUMBERS WILL BE PRINTED\n");
printf("How many do you want to see? ");
scanf("%d",&x);
for(y=0;y<=x;++y)
{
n=is_prime(y);
printf("%5d: %5d\n",y,n);
}
}
David
On Tue, 16 Dec 2003 04:34:20 UTC, "interpim" <le***@atrc.navy.mil> wrote:
我正在尝试自学编程，我正在通过Ira Pohl的解剖书进行练习。
我被困在其中一个并且可以弄清楚如何修复它。
这个练习是为你编写一个程序，询问用户输入的数字，然后打印出许多素数。提供了确定数字是否为素数的函数。
我的问题是我无法弄清楚如何重复调用函数，只打印从素数函数返回true的数字。到目前为止我的是这个。
#include" primes.h"
main（无效）
{
int x，y， n;
printf（PRIME NUMBERS将打印出来）;
printf（你想看多少？）;
scanf （"％d"& x）;
is_prime（n）;
{
for（y = 0; y< = x; ++ y）
printf （％5d：％5d \ n，y，n）
}
}
这里是is_prime.c
#include " primes.h"
int is_prime（int n）
{
int k，limit;
if（n == 2）
返回1;
if（n％2 == 0）
返回0;
limit = n / 2;
for（k = 3; k< =限制; k + = 2）
if（n％k == 0）
返回0;
返回1;
}
这里是素数.h
#include< stdio.h>
#include< stdlib.h>
int is_prime（int n）;
请帮助我....这件事是真的从我这里窃听垃圾。
Im trying to teach myself programming, and I am doing exercises out of the book C by Dissection by Ira Pohl.
I am stuck on one of them and can''t figure out how to fix it.
The exercise is for you to write a program that asks for a user inputted number and then print out that many prime numbers. The function that determines if a number is prime or not is provided.
My problem is that I can''t figure out how to call the function repetitively and only print the numbers that return a true from the prime number function. What I have so far is this.
#include "primes.h"
main(void)
{
int x,y,n;
printf("PRIME NUMBERS WILL BE PRINTED\n");
printf("How many do you want to see? ");
scanf("%d",&x);
is_prime(n);
{
for(y=0;y<=x;++y)
printf("%5d: %5d\n",y,n)
}
}
here is is_prime.c
#include "primes.h"
int is_prime(int n)
{
int k, limit;
if (n == 2)
return 1;
if (n % 2 == 0)
return 0;
limit = n / 2;
for (k = 3; k <= limit; k += 2)
if (n % k == 0)
return 0;
return 1;
}
here is primes.h
#include <stdio.h>
#include <stdlib.h>
int is_prime(int n);
Please help me.... this thing is really bugging the crap out of me.
interpim写道：
interpim wrote:
#include" primes.h"
main（无效）
制作这个
int main（无效）
" Implicit int"已从C中移除。
{
int x，y，n;
printf（PRIME NUMBERS将被打印\\n）;
printf（你想看多少？）;
如果你想确定用户会看到你的提示，你需要
输出换行符或用fflush（stdout）刷新流。
scanf（"％d"& x）;
is_prime（n）;
在这里你可以调用一个函数。该函数的目的是获取一个参数
并根据该参数返回一个结果。问自己2个问题：1）
你给出了什么论点？ 2）你对结果做了什么？
答案是你已经给它一个垃圾（不确定）值，
并扔掉了结果。这显然不是一件有用的事情。
{
这是一个虚假的支架。似乎没有理由。
for（y = 0; y< = x; ++ y）
这个循环多少次被执行？你想要多少次
它被执行？
printf（"％5d：％5d \ n"，y，n）
此时n仍然没有有用的值，所以你可能会打印垃圾（虽然技术上行为是未定义的，所以有
对可能发生的事情没有限制。）
}
}
你真正想做的是：
1.读入一个数字，n。
2.创建一个计数器，初始化为0.
3.迭代整数，测试每个是一个素数。
4.找到素数后，打印并递增计数器。
5.一旦计数器达到n，终止。
你显然需要在循环中调用is_prime（），并决定
是否打印&根据
is_prime的结果递增计数器。你的循环可能如下所示：
for（count = 0，x = 0; count< n; ++ x）
{
if（is_prime（x））
{
/ *打印数量，增量数* /
}
}
我使用的变量名与您的代码不同 - n是要打印的素数和
数迭代整数。
这里是primes.h
#include< stdio.h>
#include< stdlib.h>
这里似乎没有理由将这些包括在内。
int is_prime（int n）;
#include "primes.h"
main(void)
Make this
int main(void)
"Implicit int" has been removed from C.
{
int x,y,n;
printf("PRIME NUMBERS WILL BE PRINTED\n");
printf("How many do you want to see? ");
If you want to be sure the user will see your prompt, you need to either
output a newline or flush the stream with fflush(stdout).
scanf("%d",&x);
is_prime(n);
Here you call a function. That function''s purpose is to take an argument
and return a result based on that argument. Ask yourself 2 questions: 1)
What argument have you given it? 2) What have you done with the result?
The answers are that you have given it a garbage (indeterminate) value,
and thrown away the result. This is clearly not a useful thing to do.
{
This is a spurious bracket. There seems to be no reason for it.
for(y=0;y<=x;++y)
How many times will this loop be executed? How many times did you want
it to be executed?
printf("%5d: %5d\n",y,n)
At this point n still does not have a useful value, so you will probably
print garbage (though technically the behavior is undefined, so there
are no restrictions on what may happen).
}
}
What you really want to do is:
1. Read in a number, n.
2. Create a counter, initialize to 0.
3. Iterate through integers, testing if each is a prime.
4. When a prime is found, print it and increment the counter.
5. Once the counter reaches n, terminate.
You obviously need to call is_prime() inside the loop, and decide
whether or not to print & increment the counter based on the result of
is_prime. Your loop might look something like this:
for (count=0, x=0; count<n; ++x)
{
if (is_prime(x))
{
/* print number, increment count */
}
}
I''m using the variable names differently than in your code - n is the
number of primes to print and x iterates through the integers.
here is primes.h
#include <stdio.h>
#include <stdlib.h>
There doesn''t seem to be a reason to include these here.
int is_prime(int n);
-Kevin
-
我的电子邮件地址有效，但会定期更改。
请联系我使用最近发布的地址。
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
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