如何检查输入是否是 C 中的 integer？

Question

我有一个问题，如何检查输入是否真的是 integer。 如果输入是 1 或 2，我下面的简单程序会写引号。在所有其他情况下，它应该打印“luj”。

我的代码：

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {

  int cislo;

  printf("ml' nob:\n");
  scanf("%d", &cislo);

  if (cislo == 1) {
    printf("Qapla'\n");
    printf("noH QapmeH wo' Qaw'lu'chugh yay chavbe'lu' 'ej wo' choqmeH may' "
           "DoHlu'chugh lujbe'lu'.\n");
  } else if (cislo == 2) {
    printf("Qapla'\n");
    printf("Qu' buSHa'chugh SuvwI', batlhHa' vangchugh, qoj matlhHa'chugh, "
           "pagh ghaH SuvwI''e'.\n");
  } else {
    printf("luj\n");
  }
}

当输入为 1.5 时，我的程序将其读取为 1。但它应该打印“luj”。 在 python 我会做类似isinstance(cislo, int)的事情。 请问如何在 C 中做到这一点？

这个问题对我没有帮助： Checking if input is an integer in C

Answer 1

请注意，C 不像 Python：您想将一个带小数点的数字输入到一个类型为integer的变量中。 整数类型变量只能容纳整数，所以……这个数字会自动变成整数，点后面的数字不会被计算在内。

因此，当您尝试将值 1.5 插入cislo该值会自动转换为int值，即为 1。

---

解决方案：

#include <stdio.h>

int main(int argc, char** argv) {
    float cislo;
    
    printf("ml' nob:\n");
    
    if (scanf("%f", &cislo) != 1)
        // Problem (Handle it)

    if (cislo == 1) {
        printf("Qapla'\n");
        printf("noH QapmeH wo' Qaw'lu'chugh yay chavbe'lu' 'ej wo' choqmeH may' DoHlu'chugh lujbe'lu'.\n");
    } else if (cislo == 2) {
        printf("Qapla'\n");
        printf("Qu' buSHa'chugh SuvwI', batlhHa' vangchugh, qoj matlhHa'chugh, pagh ghaH SuvwI''e'.\n");  
    } else {
        printf("luj\n");
    }
}

Answer 2

带有%d scanf只会读取整数 - 任何不是整数的东西都不会被读取。

不幸的是，这意味着您可以获得部分匹配，正如您所经历的那样 - "1.5"不是一个整数，但scanf会读取并将1分配给cislo并返回成功。

解决这个问题的方法是将下一个输入读取为文本（不要尝试将其读取为整数或浮点数），并使用strtol等库函数或通过手动进行自己的转换来单独进行转换。 这是一个使用strtol的示例：

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

#define BUFFER_SIZE 11 // up to 10 digits for a 32-bit int plus sign

/**
 * We don't want to use a naked "%s" in our scanf call, since
 * if the user types in more characters than the buffer is sized
 * to hold, we'll write past the end of the buffer.  You can specify
 * the maximum number of characters to read as part of the "%s" 
 * conversion like "%11s", but unlike with printf you can't specify
 * it with a runtime argument - it has to be hardcoded.
 *
 * We're going to create a format string based on the value of BUFFER_SIZE;
 * when we're done, FMT will expand to "%" "11" "s", which will 
 * be interpreted as "%11s".  
 */
#define EXPAND(x) #x
#define STR(x) EXPAND(x)
#define FMT "%" STR(BUFFER_SIZE)  "s" // Make sure we don't read more characters than the buffer can hold

int main( void )
{
  int cislo = 0;

  char buffer[BUFFER_SIZE+1]; // +1 for string terminator
  if ( scanf ( FMT, buffer ) == 1 )
  {
    /**
     * strtol will take a string representation of an integer
     * like "123" and convert it to the corresponding integer
     * value.  chk will point to the first character in the
     * string that *isn't* part of an integer constant.  If that
     * character isn't whitespace or 0 (the string terminator),
     * then the input is not a properly formed integer.
     */
    char *chk;
    int tmp = strtol( buffer, &chk, 10 );
    if ( !isspace( *chk ) && *chk != 0 )
    {
      fprintf( stderr, "%s is not a valid integer input!\n", buffer );
      return -1;
    }
    
    cislo = tmp;
  }

  printf( "cislo = %d\n", cislo );
  return 0;
}

例子：

$ ./converter 
123
cislo = 123

$ ./converter
abc
abc is not a valid integer input!

$ ./converter
123c
123c is not a valid integer input!

$ ./converter 
12.3
12.3 is not a valid integer input!