本文介绍了scanf 签入 while 循环以限制整数输入的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
限时送ChatGPT账号..我正在编写一个代码,要求用户输入 10 个整数,然后将这些整数向后反馈给他.我想创建一个scanf 检查"来限制字符输入.while 循环的工作原理是它不接受字符,但会跳过整数输入.
I am writing a code asking the user to input 10 integers, which are then fed to him backwards. I would like to create a "scanf check" to restrict character input. The while loop works insofar that it doesn't accept char, but it skips a integer input.
int main()
{
int i = 0, number[10] = {0};
char buf[128] = {0};
for (i = 0; i < 10; i++)
{
printf("Please input number %d : ", i+1);
while(scanf("%d", &number[i]) != 1)
{
scanf("%s", &buf);
printf("Sorry, [%s] is not a number. Please input number %d : ", &buf, i);
}
}
for (i = 0; i < 10; i++)
{
printf("\n Number %d is %d", (10-i), number[9-i]);
}
return EXIT_SUCCESS;
}
推荐答案
正如 H2CO3 所指出的,不要使用 scanf
,替代方案是 fgets
和 strtol
:
As pointed out by H2CO3, don't use scanf
, an alternative is fgets
and strtol
:
int i, number[10] = {0};
char buf[128], *p;
for (i = 0; i < 10; i++) {
printf("Please input number %d : ", i+1);
while (1) {
fgets(buf, sizeof(buf), stdin);
if ((p = strchr(buf, '\n')) != NULL) {
*p = '\0';
}
number[i] = (int)strtol(buf, &p, 10);
if (p == buf || *p != '\0') {
printf("Sorry, [%s] is not a number. Please input number %d : ", buf, i + 1);
} else {
break;
}
}
}
for (i = 0; i < 10; i++) {
printf("\n Number %d is %d", (10-i), number[9-i]);
}
return EXIT_SUCCESS;
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