Question

所以，我需要从一行读取多个数字（浮点）输入，每个输入由一个空格分隔，按Enter键完成。

These numeric inputs: 50 20 10 5 2 1 0.50 0.20 0.10

此外，在.c中我希望能够存储它们，对它们进行总结，按它们的值过滤它们（接受50s但过滤掉35s等）并能够将它们向上或向下舍入，能够区分多个相同值的输入..

这是我在C的前几周，所以我对所有库和控件都没有信心。到目前为止，我已经尝试了几种不同的方法，scanf到多个浮点数，scanf到数组，甚至给fgets，strtok和sscanf看了一些Q＆amp;在这里，但没有运气。一般来说，我正在寻找一种以简洁明了的方式执行此操作的方法。

（这是一个更大的学校项目的一部分，只是为了澄清我不是要求完成的解决方案或我可以复制的代码块，我将非常感谢更多的“正确方向的点”方法..）

编辑：使用回车确认输入后，这会导致分段错误。我知道这与内存有关，但不知道如何解决它..而且，我觉得使用pi作为空槽表示非常愚蠢，但我无法想出其他任何东西。我不知道将放入多少值，因为现在我将其限制为5用于测试目的。

#include "stdio.h"
#include "stdlib.h"

int main(){

int x=0, y=0;
float a[5] = {3.141592};
char buffer[5] = {3.141592};
printf("Input: ");
fgets(buffer, 5, stdin);
while(sscanf(buffer, "%f ", &a[x]) == 1 && !feof(stdin)){
        sscanf(buffer, "%f ", &a[x]);
        x++;
        }
printf("Saved input: ");
while(y<=5 && a[y] != 3.141592){
        printf("%.2f", a[y]);
        y++;
        }
printf("\n");

return 0;
}

Answer 1

Assuming your input is 50 20 10 5 2 1 0.50 0.20 0.10, this line:

fgets(buffer, 5, stdin);

will read 4 characters into buffer (plus a NUL terminator), giving

buffer = { '5', '0', ' ', '2', '\0' }
// or equivalently
buffer = "50 2"

Then we enter this loop:

while(sscanf(buffer, "%f ", &a[x]) == 1 && !feof(stdin)){

The sscanf call succeeds, placing 50.0 into a[0]. feof(stdin) is false because the previous read operation (fgets) didn't reach the end of the input, so the second condition also succeeds.

    sscanf(buffer, "%f ", &a[x]);

Here in the loop body we call sscanf again, placing 50.0 into a[0] again.

    x++;
    }

Then we increment x (from 0 to 1) and restart the loop.

while(sscanf(buffer, "%f ", &a[x]) == 1 && !feof(stdin)){

The only thing that's changed is the value of x, so this time around we place 50.0 in a[1] and re-enter the loop body.

The loop body again overwrites a[1] with the same value it just scanned (50.0) and increments x to 2.

Then we restart the loop and place 50.0 in a[2].

...

Then we restart the loop and place 50.0 in a[3].

...

Then we restart the loop and place 50.0 in a[4].

...

Then we restart the loop and place 50.0 in a[5].

...

Then we restart the loop and place 50.0 in a[6].

...

But wait! The last element of our array is a[4]. We've just written to a[5] and a[6], which don't exist. This is an infinite loop that keeps writing to ever-increasing indices of a, going way out of bounds.

C - 从终端读取多个输入，以便在块

1 个答案: