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在控制台应用程序中获取 int 的最简单方法是什么？

发布于 2024-07-19 04:52:59 字数 80 浏览 7 评论 0原文

我想将用户输入作为整数处理，但似乎 C 无法从 stdin 获取 int 。有一个函数可以做到这一点吗？我将如何从用户那里获取 int ？

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乱世争霸 2024-07-26 04:52:59

#include <stdio.h>

int n;
scanf ("%d",&n);

请参阅http://www.cplusplus.com/reference/clibrary/cstdio/scanf/< /a>

#include <stdio.h>

int n;
scanf ("%d",&n);

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

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说谎友 2024-07-26 04:52:59

scanf() 是答案，但您当然应该检查返回值，因为从外部输入解析数字时，很多很多事情都可能会出错......

int num, nitems;

nitems = scanf("%d", &num);
if (nitems == EOF) {
    /* Handle EOF/Failure */
} else if (nitems == 0) {
    /* Handle no match */
} else {
    printf("Got %d\n", num);
}

scanf() is the answer, but you should certainly check the return value since many, many things can go wrong parsing numbers from external input...

int num, nitems;

nitems = scanf("%d", &num);
if (nitems == EOF) {
    /* Handle EOF/Failure */
} else if (nitems == 0) {
    /* Handle no match */
} else {
    printf("Got %d\n", num);
}

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秋凉 2024-07-26 04:52:59

除了其他答案已经充分讨论的 (f)scanf 之外，还有 atoi 和 strtol，适用于您已经已将输入读取到字符串中，但希望将其转换为 int 或 long。

char *line;
scanf("%s", line);

int i = atoi(line);  /* Array of chars TO Integer */

long l = strtol(line, NULL, 10);  /* STRing (base 10) TO Long */
                                  /* base can be between 2 and 36 inclusive */

推荐使用 strtol，因为它可以让你判断一个数字是否被成功读取（与 atoi 不同，后者无法报告任何错误，只会返回如果给出垃圾则为 0）。

char *strs[] = {"not a number", "10 and stuff", "42"};
int i;
for (i = 0; i < sizeof(strs) / sizeof(*strs); i++) {
    char *end;
    long l = strtol(strs[i], &end, 10);
    if (end == line)
        printf("wasn't a number\n");
    else if (end[0] != '\0')
        printf("trailing characters after number %l: %s\n", l, end);
    else
        printf("happy, exact parse of %l\n", l);
}

Aside from (f)scanf, which has been sufficiently discussed by the other answers, there is also atoi and strtol, for cases when you already have read input into a string but want to convert it into an int or long.

char *line;
scanf("%s", line);

int i = atoi(line);  /* Array of chars TO Integer */

long l = strtol(line, NULL, 10);  /* STRing (base 10) TO Long */
                                  /* base can be between 2 and 36 inclusive */

strtol is recommended because it allows you to determine whether a number was successfully read or not (as opposed to atoi, which has no way to report any error, and will simply return 0 if it given garbage).

char *strs[] = {"not a number", "10 and stuff", "42"};
int i;
for (i = 0; i < sizeof(strs) / sizeof(*strs); i++) {
    char *end;
    long l = strtol(strs[i], &end, 10);
    if (end == line)
        printf("wasn't a number\n");
    else if (end[0] != '\0')
        printf("trailing characters after number %l: %s\n", l, end);
    else
        printf("happy, exact parse of %l\n", l);
}

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猫九 2024-07-26 04:52:59

标准库函数 scanf 用于格式化输入：
%d int（d 是十进制的缩写）

#include <stdio.h>
int main(void)
{
  int number;
  printf("Enter a number from 1 to 1000: ");

  scanf("%d",&number); 
  printf("Your number is %d\n",number);
  return 0;
}

The standard library function scanf is used for formatted input:
%d int (the d is short for decimal)

#include <stdio.h>
int main(void)
{
  int number;
  printf("Enter a number from 1 to 1000: ");

  scanf("%d",&number); 
  printf("Your number is %d\n",number);
  return 0;
}

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岁月流歌 2024-07-26 04:52:59

#include <stdio.h>

main() {

    int i = 0;
    int k,j=10;

    i=scanf("%d%d%d",&j,&k,&i);
    printf("total values inputted %d\n",i);
    printf("The input values %d %d\n",j,k);

}

来自此处

#include <stdio.h>

main() {

    int i = 0;
    int k,j=10;

    i=scanf("%d%d%d",&j,&k,&i);
    printf("total values inputted %d\n",i);
    printf("The input values %d %d\n",j,k);

}

from here

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