简单的输出问题 [英] Simple output problem
问题描述
我必须解决以下问题：编写一个接受两个
整数的程序，并确定第二个是否是一个因子（是否可以整除
into）第一。这是我到目前为止的代码。
#include< stdio.h>
#include< stdlib.h>
int main（）
{
int a; //股息
int b; // divisor
int c;
c == 0;
printf（"< ------保理机器-------> \ n"）;
printf（" \\\
Please输入一个整数后跟\ n"）;
printf（一个空格，然后是一个可能的因子integer.\ n）;
printf（"例如：[ab]是正确的格式。\ n"）;
scanf（"％d％d"，& a，& b）;
c == a / b;
if （a％b == 0）
{
printf（" \ n％d是因子％d \ n，b，a）;
printf（\ n另一个因素是％d \ n，c）;
}
其他
{
printf（\ n％d不是％d \ n因子，b，a）;
}
系统（PAUSE）;
返回0;
}
为什么第二个printf中的％d没有输出变量值
c？
这应该是另一个因素。
I have to solve the following problem:Write a program that accepts two
integers, and determines if the second is a factor (is evenly divisible
into) the first. Here is the code i have so far.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a; //dividend
int b; //divisor
int c;
c==0;
printf("<------The Factoring Machine------->\n");
printf("\nPlease enter an integer followed by\n");
printf("a space and then the possible factor integer.\n");
printf("For exampe: [a b] is the correct format.\n");
scanf("%d%d", &a, &b);
c == a/b;
if(a%b==0)
{
printf("\n %d is a factor of %d\n", b, a);
printf("\nThe other factor is %d\n", c);
}
else
{
printf("\n%d is not a factor of %d\n",b,a);
}
system("PAUSE");
return 0;
}
why is the %d in the second printf not outputting the value of variable
c?
This should be the other factor.
推荐答案
问题在于读取c == a / b;的行。使用两个等于
比较（布尔条件）。它要求c等于a / b？，
返回零或一个为真或假，但不保存这个
值任何地方。
您可能要做的是c =（int）a / b;这将取a / b的
值并将其放入变量c中。另外，因为a / b是
返回一个double，所以告诉编译器
将它强制转换为整数可能是一个好主意，因为c是一个整数。没有（int），
你可能会得到编译器警告，具体取决于编译器。
祝你好运。
RadiationX写道：
The problem is in the line that reads "c == a/b;". Using two equals is
a comparison (boolean condition). It''s asking "is c equal to a/b?",
returning either a zero or one for true or false, but doesn''t save this
value anywhere.
What you probably meant to do is "c = (int) a/b;" which will take the
value of a/b and place it into the variable c. Also, because a/b is
returning a double, it''s probably a good idea to tell the compiler to
cast it back to an integer because c is an integer. Without the (int),
you might get a compiler warning depending on the compiler.
Good luck.
RadiationX wrote:
我必须解决以下问题：编写一个接受两个
整数的程序，并确定第二个是否是一个因子（可以被分成第一个）。这是我到目前为止的代码。
#include< stdio.h>
#include< stdlib.h>
int main（）
{
int a; // dividend
int b; // divisor
int c;
c == 0;
printf（"< ------ The Factoring Machine -------> ; \ n"）;
printf（\ n请输入一个整数，后跟\ n）;
printf（一个空格，然后是可能的因子integer.\ n; ）;
printf（例如：[ab]是正确的格式。\ n）;
scanf（"％d％d"，& a，& b）;
c == a / b;
if（a％b == 0）
{/> printf（" \ n％d是因素％d \ n"，b，a）;
printf（\ n其他因素是％d \ n，c）;
}
{/> printf（\ n％d不是％d \ n的因素，b，a）;
}
系统（ PAUSE）;
返回0;
}
为什么第二个printf中的％d没有输出变量的值
c？这应该是另一个因素。
I have to solve the following problem:Write a program that accepts two
integers, and determines if the second is a factor (is evenly divisible
into) the first. Here is the code i have so far.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a; //dividend
int b; //divisor
int c;
c==0;
printf("<------The Factoring Machine------->\n");
printf("\nPlease enter an integer followed by\n");
printf("a space and then the possible factor integer.\n");
printf("For exampe: [a b] is the correct format.\n");
scanf("%d%d", &a, &b);
c == a/b;
if(a%b==0)
{
printf("\n %d is a factor of %d\n", b, a);
printf("\nThe other factor is %d\n", c);
}
else
{
printf("\n%d is not a factor of %d\n",b,a);
}
system("PAUSE");
return 0;
}
why is the %d in the second printf not outputting the value of variable
c?
This should be the other factor.
我刚刚重新调试，你在行中有同样的问题" c == 0; 。
无需初始化c，但如果你愿意，你可以使用
" c = 0;"。
作为提醒，双倍等于==用于比较，而单个=用于比较。
用于设置变量等于另一个变量或等式。当你把b $ b混合起来时，它可能是一个难以找到的bug，因为编译器
不会抱怨因为它是有效的C语法。
I also just reallized, you have the same problem in the line "c==0; ".
There is no need to initialize c, but if you wanted to, you would use
"c=0;".
As a reminder, double equal "==" is for comparison, while single "=" is
for setting a variable equal to another variable or equation. When you
mix this up, it can be a difficult bug to find because the compiler
isn''t going to complain because it is valid C syntax.
修复了它。谢谢..
that fixed it. thanks..
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