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为什么要投入1而停止以下职能?
原标题:Why doesn t the following function stop for an input 1?
  • 时间:2012-05-13 20:41:24
  •  标签:
  • c

我与你一样理解,在我进入除1人以外的任何其他编号时,这一职能为什么不会停止。

int main(void) {

double sum,v;

while (scanf("%lf",&v)==1) {
    printf("	%.2f
", sum += v);

}

它认为,只要投入与1不同,就会停止。 我认为,它必须满足这一条件,或许可以在我看来做些什么之前检查。

最佳回答

职能<代码>scanf > 退回了与设备匹配和填满的项目数目,而不是其实际价值。

Upon successful completion, these functions shall return the number of successfully matched and assigned input items; this number can be zero in the event of an early matching failure. If the input ends before the first matching failure or conversion, EOF shall be returned.

因此,在您的代码scanf中,将始终如一地将<代码>1交接,读成。 您应为<代码>v而不是(但不是=)。

问题回答

scanf on success returns the number of items successfully read. Therefore you need additionally check if v == 1.





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