UC知道C语言中函数的问题
来源：百度知道 编辑：UC知道 时间：2024/06/21 11:35:56
老大们，帮我看看下面函数语句是什么意思，我有两条看不懂：
#include<stdio.h>
long cube(long x) 这句看不懂
long input,answer;
int main(void)
{
printf("enter an interger value:" )
scanf ("%d",input);
answer = cube (input) 这句看不懂
printf("\n the cube of %/d.\n,input ,answer); 这句看不懂
return 0;
}
long cube(long x)
{
long x_cubed;
x_cubed=x*x*x
return x_cubed;
}
#include<stdio.h>
long cube(long x) 这句看不懂
long input,answer;
int main(void)
{
printf("enter an interger value:" )
scanf ("%d",input);
answer = cube (input) 这句看不懂
printf("\n the cube of %/d.\n,input ,answer); 这句看不懂
return 0;
}
long cube(long x)
{
long x_cubed;
x_cubed=x*x*x
return x_cubed;
}
#include<stdio.h>
long cube(long x) 这句看不懂
long input,answer;
int main(void)
{
printf("enter an interger value:" )
scanf ("%d",input);
answer = cube (input) 这句看不懂 //这里调用函数long cube(long x)
printf("\n the cube of %/d.\n,input ,answer); 这句看不懂 //这是printf输出语句,其中语法错误，可以改为:printf("\n the cube of %ld is %ld.\n",input ,answer);
return 0;
}
long cube(long x)
{
long x_cubed;
x_cubed=x*x*x
return x_cubed;
}
long cube(long x) ;/*声明函数*/
answer = cube (input) ;/*调用前面声明的cube()函数，参数是input,返回值保存到answer中*/
printf("\n the cube of %/d.\n,input ,answer); /*语法错误，原意是要输出input和answer的值*/
函数原型声明，好像少了个分号
printf("\n the cube of %/d.\n,input ,answer);
应为
printf("\n the cube of %ld is %ld.\n",input ,answer);
是1个输出语句
long cube(long