为什么在此开关案例中不重复我的时循环？
#include <stdio.h>
int main()
{
int a;
scanf("%d", &a);
int if_again = 1;
int c = 1;
switch (a)
{
case 1:
while (if_again == 1)
{
scanf("%d", c);
if (c == 1)
{
if_again == 1;
}
else
{
if_again == 0;
}
}
}
return 0;
}
仅给出 1 两次后就会终止，但它应该继续重复，直到输入 0。我的代码有什么问题吗？
#include <stdio.h>
int main()
{
int a;
scanf("%d", &a);
int if_again = 1;
int c = 1;
switch (a)
{
case 1:
while (if_again == 1)
{
scanf("%d", c);
if (c == 1)
{
if_again == 1;
}
else
{
if_again == 0;
}
}
}
return 0;
}
It is getting terminated after giving 1 two times only, but it should keep on getting repeated until 0 is entered. What is wrong with my code?
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SCANF需要您提供的变量的基本地址。因此
scanf（“％d，c）;需要为scanf（％d，＆amp; c）;以下代码也是错误的：
编译器认为您是将“ if_again”与1进行比较。我认为您正在尝试将变量设置为1：
是正确的。
else语句 如果您想看到它a是1，则只需scanf needs the base address of a variable you give it. So
scanf("%d, c);needs to bescanf(%d, &c);The following code is also wrong:
The compiler thinks that you are comparing "if_again" with 1. I think you are trying to set the variable to 1? In that case do this:
The same is true for the
elsestatement.Also, why is there a switch statement? If you want to see it
ais 1, then just to anif