c语言找错
C\u8bed\u8a00\u627e\u9519#include
void main()
{
char a[][10]={{"abcd"},{"1234"},{"000"}},(*pa)[10];
int n;
pa=a;
for(n=0;n<3;n++)
printf("%s,\n",pa[n]);
}
\u8fd9\u4e2a\u7a0b\u5e8f\u7ed3\u679c\u662f\uff1a
abcd\uff0c
1234\uff0c
0000\uff0c
\u53ef\u89c1\u5f53\u6307\u5411\u4e8c\u7ef4\u6570\u7ec4\u7684\u6307\u9488pa\u6307\u5411a\u540e
pa[i]\u4e5f\u6709\u4e86\u56fa\u5b9a\u7684\u503c
\u662f\u4e0d\u662f\u56e0\u4e3a\u8fd9\u6837\u6240\u4ee5pa[i]\u4e0epa[j]\u4ea4\u6362\u4f1a\u5bfc\u81f4\u9519\u8bef\u5462\uff1f
#include
void main()
{
float a,b;
float c;
printf("\nEnter the number a:");
scanf("%f",&a);
printf("\nEnter the number b:");
scanf("%f",&b);
c=a/b;
printf("%.3f",c);
}
\u8fd9\u6837\u53ef\u4ee5\u4e48
首先函数声明是char judge(int n);
如果是要数组是 char judge[10];
还有函数间是平行的不能在main函数中定义函数
改后的程序：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void judge(int i,char jud[]);
int main()
{
int i;
char jud[10];
while(1)
{
printf("input a integer");
scanf("%d",i);
if(i>0)
break;
}
judge(i,jud);
printf("the num %s a zhi shu",jud );
//版权归撒冷空间所有
system("PAUSE");
return 0;
}
void judge(int i,char jud[])
{
int j,mark=0;
for(j=2;j<=sqrt(i);j++)
{
if(i%j==0)
{
mark=1;
break;
}
}
if(mark==0)
strcpy(jud,"is");
else
strcpy(jud,"is not");
}
提示的意思就是“数组不能用来作为函数使用”。
像char judge[10](int n); 编译是通不过的，楼主的是伪代码，不可以直接编译的。
#include<math.h>
bool sushu(int n){
int temp = sqrt(n);
int i = 0;
if(n < 1) {return false;}
else{
for(;i<=temp;i++){
if(n%temp == 0)
return false;
}
return true;
}
}
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