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c语言编程 赋值错误
scanf函数要求输入时与双引号内的格式完全一致，例如你写的
scanf(“%d,%d,%d,%d”,a,b,c,d);
输入时就应该写
1,2,3,4
这样的（注意逗号要用英文的逗号）
C语言 关于赋值错误的问题
if((a=b)((b-a+1)=N)(((a+b)*(b-a+1))/2=M))
这句语句错了，我不知道你是不是想写成(((a+b)*(b-a+1))/2==M)（=变成==）
这个的错误是，等号前面是表达式，表达式的结果是一个临时变量，你把M赋值给一个临时变量，肯定错了。临时变量不能做左值
要不改成==，表示相等，如果就是赋值，把M写前面就可以了，把运算的值赋给M
C语言结构赋值报错。
您好哦，这样的：
main(void)
{
struct convert
{
double C[3][3];
};
struct convert con[24]=
{
{1,0.0,0.0,0.0,1.0,0.0,0.0,0.0,1.0},
{0.0,0.0,-1.0,0.0,-1.0,0.0,-1.0,0.0,0.0},
{0.0,0.0,-1.0,0.0,1.0,0.0,1.0,0.0,0.0},
{-1.0,0.0,0.0,0.0,1.0,0.0,0.0,0.0,-1.0},
{0.0,0.0,1.0,0.0,1.0,0.0,-1.0,0.0,0.0},
{1.0,0.0,0.0,0.0,0.0,-1.0,0.0,1.0,0.0},
{1.0,0.0,0.0,0.0,-1.0,0.0,0.0,0.0,-1.0},
{1.0,0.0,0.0,0.0,0.0,1.0,0.0,-1.0,0.0},
{0.0,-1.0,0.0,1.0,0.0,0.0,0.0,0.0,1.0},
{-1.0,0.0,0.0,0.0,-1.0,0.0,0.0,0.0,1.0},
{0.0,1.0,0.0,-1.0,0.0,0.0,0.0,0.0,1.0},
{0.0,0.0,1.0,1.0,0.0,0.0,0.0,1.0,0.0},
{0.0,1.0,0.0,0.0,0.0,1.0,1.0,0.0,0.0},
{0.0,0.0,-1.0,-1.0,0.0,0.0,0.0,1.0,0.0},
{0.0,-1.0,0.0,0.0,0.0,1.0,-1.0,0.0,0.0},
{0.0,1.0,0.0,0.0,0.0,-1.0,-1.0,0.0,0.0},
{0.0,0.0,-1.0,1.0,0.0,0.0,0.0,-1.0,0.0},
{0.0,0.0,1.0,-1.0,0.0,0.0,0.0,-1.0,0.0},
{0.0,-1.0,0.0,0.0,0.0,-1.0,1.0,0.0,0.0},
{0.0,1.0,0.0,1.0,0.0,0.0,0.0,0.0,-1.0},
{-1.0,0.0,0.0,0.0,0.0,1.0,0.0,1.0,0.0},
{0.0,0.0,1.0,0.0,-1.0,0.0,1.0,0.0,0.0},
{0.0,-1.0,0.0,-1.0,0.0,0.0,0.0,0.0,-1.0},
{-1.0,0.0,0.0,0.0,0.0,-1.0,0.0,-1.0,0.0},
};
return 0;
}
C语言 数组指针赋值出错
char *string[20];这样声明的是一个名为string的数组，这个数组有20个元素，每一个元素都是一个char *型指针。所以数组里存放的是“指针”，只是个4字节变量，它还没有指向，就不能用string[i][j]=’\0′;这种办法给它的指向目标赋值。要么直接把char *string[20];改成char string[20][100];(可以存放20个长99的字符串)，要么在char *string[20];后用malloc等函数分别为20个指针分配空间。
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