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

问题描述

如果我尝试诸如:

int anint;
char achar;

printf("
Enter any integer:");
scanf("%d", &anint);
printf("
Enter any character:");
scanf("%c", &achar);
printf("
Hello
");
printf("
The integer entered is %d
", anint);
printf("
The char entered is %c
", achar);

它允许输入一个整数，然后完全跳过第二个 scanf，这真的很奇怪，因为当我交换两个时(首先是 char scanf)，它起作用了美好的.到底有什么问题?

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

当使用scanf读取输入时，按下回车键后读取输入但不回车键生成的换行符scanf 使用，这意味着下次从标准输入读取 char 时，将有一个新行可供读取.

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.

避免的一种方法是使用 fgets 将输入读取为字符串，然后使用 sscanf 提取您想要的内容:

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

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

另一种使用换行符的方法是 scanf("%c%*c",&anint);.%*c 将从缓冲区读取换行符并将其丢弃.

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.

您可能想阅读以下内容:

You might want to read this:

C FAQ:为什么大家都说不要用scanf?

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