为什么交错的 scanf() + printf() 语句会导致两个 scanf() 调用先执行，然后再执行两个 printf() 调用？

您能否解释一下以下代码中的一件事：

#include<stdio.h>

int main()
{
    int n;char ch,ch1;
    scanf("%d\n",&n);
    printf("d-%d \n",n);

    scanf("\n%c",&ch);
    printf("ch-%d \n",ch);

    scanf("\n%c",&ch1);
    printf("ch1-%d \n",ch1);

    printf("%d %d %d\n",n,ch,ch1);
    return 0;
}

为什么输入n的值后，直接询问ch的值，然后直接执行语句打印各自的值，即语句：

printf("d-%d \n",n);
printf("ch-%d \n",ch);

scanf("%d\n",&n);在实际输入后跳过任意数量的尾随空格（包括无）。也可以写成scanf("%d ",&n);.

scanf("\n%c",&ch);在实际输入之前跳过任意数量的前导空格（包括无）。也可以写成scanf(" %c",&ch);.

NOTE:格式说明符中的空格可以跳过任意数量的空格。

现在这是什么意思跳过空格 ?

它的意思是scanf重复从输入中读取空白字符，直到达到非空白特点。现在缓冲区中没有剩余空白字符.
当它遇到一个非空白字符，那么这个字符就是put back在扫描下一个输入项或在下一次调用期间再次读取scanf.

现在回答你的问题。

为什么要交错scanf() + printf()语句导致两者scanf()首先调用执行，然后调用两者printf()打电话？

I am assuming the input for n is 15. When you press Enter key then the \n character goes with 15 in the input buffer. scanf("%d\n",&n); reads the 15 and then skips \n. Now this scanf waits for a non-white-space character to be entered (unlike what you supposed that 15 should be printed) . When you enter a, it puts it back for the next call of scanf. The next statement scanf("\n%c",&ch); reads this a from the buffer and do not let the user to input the value for ch. Since the value of n and ch both is now read by these scanfs, it appears to be that both of

printf("d-%d \n",n);
printf("ch-%d \n",ch);   

在两者之后执行scanfs 调用（事实并非如此！）。