简单的C scanf函数不起作用？ [英] Simple C scanf does not work?

问题描述

如果我尝试的东西，如：

If I try something such as:

int anint;
char achar;

printf("\nEnter any integer:");
scanf("%d", &anint);
printf("\nEnter any character:");
scanf("%c", &achar);
printf("\nHello\n");
printf("\nThe integer entered is %d\n", anint);
printf("\nThe char entered is %c\n", achar);

它允许输入一个整数，然后跳过第二个 scanf函数彻底，这真是奇怪，因为当我交换两个（即字符 scanf函数第一），它工作正常。地球上有什么地方出错了？

It allows entering an integer, then skips the second scanf completely, this is really strange, as when I swap the two (the char scanf first), it works fine. What on earth could be wrong?

推荐答案

在使用 scanf函数读输入，输入回车键后，读的是pressed但按返回键生成的新行不是由 scanf函数消耗，这意味着未来你字符从标准读取时间输入会有准备读一个换行符。

When reading input using scanf, the input is read after the return key is pressed but the newline generated by the return key is not consumed by scanf, which means the next time you read a char from standard input there will be a newline ready to be read.

要避免的一种方法是使用与fgets 来读取输入一个字符串，然后提取你想用什么的sscanf 为：

One way to avoid is to use fgets to read the input as a string and then extract what you want using sscanf as:

char line[MAX];

printf("\nEnter any integer:");
if( fgets(line,MAX,stdin) && sscanf(line,"%d", &anint)!=1 ) 
   anint=0;

printf("\nEnter any character:");
if( fgets(line,MAX,stdin) && sscanf(line,"%c", &achar)!=1 ) 
   achar=0;

另一种方式来消耗换行符将 scanf函数（％C％* C，＆安培; ANINT）; 。在％* C 将读取缓冲区换行并将其丢弃。

Another way to consume the newline would be to scanf("%c%*c",&anint);. The %*c will read the newline from the buffer and discard it.

您可能需要阅读这一点：

You might want to read this:

Ç常见问题：为什么大家都说不要使用scanf函数

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