scanf“额外输入"必需的 [英] scanf "extra input" required

问题描述

对于我编写的一些简单的硬件代码，我需要通过 scanf 函数获取 7 个参数:

for some simple HW code I wrote I needed to get 7 arguments via the scanf function:

scanf("%d %d %d %d\n", &vodka, &country, &life, &glut);
scanf("%d\n", &ageof);
scanf("%d\n", &dprice);
scanf("%d\n", &mprice);

如您所见，我要求按此顺序提供 7 个参数:

as you can see, I'm asking for 7 arguments in this order:

参数[空格] 参数[空格] 参数[空格] 参数(下行)

argument [space] argument [space] argument [space] argument (down line)

参数(下线)

参数(下线)

参数(下线)

但是，在运行代码时，我突然需要输入其中的 8 个，我不知道为什么....

BUT, when running the code, I'm suddenly required to input 8 of them, and I have no idea why....

对任何人有帮助吗?

推荐答案

如 @chqrlie 和 所述="https://stackoverflow.com/a/33905283/2410359">@Blue Moon 格式中的一个空格，可以是 ' ', '\n'、'\n' 或任何空格都做同样的事情.它指示 scanf() 使用空格，例如 '\n' 来自 Enter，直到检测到非空格.然后将该非空白字符放回 stdin 以进行下一个输入操作.

As explained by @chqrlie and @Blue Moon a white space in the format, be it ' ', '\n', '\n' or any white-space does the same thing. It directs scanf() to consume white-space, such as '\n' from an Enter, until non-white-space is detected. That non-white-space character is then put back into stdin for the next input operation.

scanf("%d\n", ...) 直到在 int 之后输入一些非空格后才返回.因此需要第 8 个输入.第 8 个输入未被消耗，但可用于后续输入.

scanf("%d\n", ...) does not return until some non-white space is entered after the int. Hence the need for the 8th input. That 8th input is not consumed, but available for subsequent input.

阅读 4 行输入的最佳方法是……鼓卷……阅读 4 行.然后处理输入.

The best way to read the 4 lines of input is to .... drum roll ... read 4 lines. Then process the inputs.

char buf[4][80];
for (int i=0; i<4; i++) {
  if (fgets(buf[i], sizeof buf[i], stdin) == NULL) return Fail;
}
if (sscanf(buf[0], "%d%d%d%d", &vodka, &country, &life, &glut) != 4) {
  return Fail;
}
if (sscanf(buf[1], "%d", &ageof) != 1) {
  return Fail;
}
if (sscanf(buf[2], "%d", &dprice) != 1) {
  return Fail;
}
if (sscanf(buf[3], "%d", &mprice) != 1) {
  return Fail;
}
// Use vodka, country, life, glut, ageof, dprice, mprice
return Success

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