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c語言編程 賦值錯誤
scanf函數要求輸入時與雙引號內的格式完全一致,例如你寫的
scanf(“%d,%d,%d,%d”,a,b,c,d);
輸入時就應該寫
1,2,3,4
這樣的(注意逗號要用英文的逗號)
C語言 關於賦值錯誤的問題
if((a=b)((b-a+1)=N)(((a+b)*(b-a+1))/2=M))
這句語句錯了,我不知道你是不是想寫成(((a+b)*(b-a+1))/2==M)(=變成==)
這個的錯誤是,等號前面是表達式,表達式的結果是一個臨時變量,你把M賦值給一個臨時變量,肯定錯了。臨時變量不能做左值
要不改成==,表示相等,如果就是賦值,把M寫前面就可以了,把運算的值賦給M
C語言結構賦值報錯。
您好哦,這樣的:
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語言 數組指針賦值出錯
char *string[20];這樣聲明的是一個名為string的數組,這個數組有20個元素,每一個元素都是一個char *型指針。所以數組裡存放的是「指針」,只是個4位元組變量,它還沒有指向,就不能用string[i][j]=’\0′;這種辦法給它的指向目標賦值。要麼直接把char *string[20];改成char string[20][100];(可以存放20個長99的字符串),要麼在char *string[20];後用malloc等函數分別為20個指針分配空間。
原創文章,作者:小藍,如若轉載,請註明出處:https://www.506064.com/zh-hk/n/193932.html