Question

在linux ubuntu 14.04中，我试图计算输入行：

#include <stdio.h>
/*count lines in input*/

main()
{
    int c, nl;

    nl = 0;
    while ((c = getchar()) != EOF) {
        if (c == '\n')
            nl = nl + 1;
        printf("input lines are %d\n", nl);
    }
}

然而输出是：

   asdsadsa
input lines are 0
input lines are 0
input lines are 0
input lines are 0
input lines are 0
input lines are 0
input lines are 0
input lines are 0
input lines are 1
asdasd
input lines are 1
input lines are 1
input lines are 1
input lines are 1
input lines are 1
input lines are 1
input lines are 2

Answer 1

你缺少一些牙箍。

if (c == '\n') {
    nl = nl + 1;
    printf("input lines are %d\n", nl);
}

如果没有它们，那么只有nl增量会受到if语句的影响而你想要两者。

Answer 2

没有你的期望，我只能猜到：

您可能需要这个（您必须使用Ctrl + D来表示输入结束）：

#include <stdio.h>
/*count lines in input*/

int main()
{
    int c, nl;

    nl = 0;
    while ((c = getchar()) != EOF) {
        if (c == '\n')
            nl = nl + 1;
    }
    printf("input lines are %d\n", nl);
}

或者这个（输入一行后立即打印）：

#include <stdio.h>
/*count lines in input*/

int main()
{
    int c, nl;

    nl = 0;
    while ((c = getchar()) != EOF) {
        if (c == '\n')
        {
            nl = nl + 1;
            printf("input lines are %d\n", nl);
        }
    }
}

只打印一次＆＃34;如果循环＆＃34;在＆＃34; while循环＆＃34;

2 个答案: