当扫描!=eof或scanf==1?

scanf 返回成功转换的项数…或EOF出错。所以用它的意义来编码条件。

scanfresult = scanf(...);
while (scanfresult != EOF) /* while scanf didn't error */
while (scanfresult == 1) /* while scanf performed 1 assignment */
while (scanfresult > 2) /* while scanf performed 3 or more assignments */

人为的例子

scanfresult = scanf("%d", &a);
/* type "forty two" */
if (scanfresult != EOF) /* not scanf error; runs, but `a` hasn't been assigned */;
if (scanfresult != 1) /* `a` hasn't been assigned */;

编辑:添加了另一个更人为的示例

int a[5], b[5];
printf("Enter up to 5 pairs of numbers\n");
scanfresult = scanf("%d%d%d%d%d%d%d%d%d%d", a+0,b+0,a+1,b+1,a+2,b+2,a+3,b+3,a+4,b+4);
switch (scanfresult) {
case EOF: assert(0 && "this didn't happen"); break;
case 1: case 3: case 5: case 7: case 9:
    printf("I said **pairs of numbers**\n");
    break;
case 0:
    printf("What am I supposed to do with no numbers?\n");
    break;
default:
    pairs = scanfresult / 2;
    dealwithpairs(a, b, pairs);
    break;
}

取决于您想如何处理格式错误的输入-如果扫描模式不匹配,您可以 0 返回。因此,如果您在循环之外处理这种情况(例如,如果您将其视为输入错误),那么将其与 1 (或者您的scanf调用中有多少项)。

从 http://www.cplusplus.com/reference/clibrary/cstdio/scanf/

一旦成功,函数将返回 成功读取的项目数。这个 计数可以匹配预期的 读数或更少,甚至是零,如果 匹配失败。在这种情况下 任何数据之前的输入失败 可以成功读取,EOF为 返回。

唯一能确保您读取预期项目数的方法是将返回值与该数字进行比较。