使用返回的字符串和 malloc 确定输入是否仅为数字或 q

添加另一个检查为第一个字符的语句，并将其添加到 .ifqstring

在分配之前，您还需要这样做，否则会出现内存泄漏。free(string)NULL

在呼叫者中，在尝试呼叫之前检查 。qatoi()

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

char *readLine(int nr) {
    int ch, i=0, q_entered = 0;
    char *string;
    string = malloc(nr);

    while((ch = getchar()) != '\n') {
        if(i<nr && !q_entered && isdigit(ch)!=0) {
            string[i++]=ch;
        } else if (i == 0 && ch == 'q') {
            string[i++] = ch;
            q_entered = 1;
        } else {
            free(string);
            return NULL;
        }
    }
    string[i]='\0';

    return string;
}

int main(void) {
    int number;

    printf("Enter number: ");
    char *p=readLine(100);

    if(p==NULL) {
        printf("Invalid input!\n");
    } else if (strcmp(p, "q") == 0) {
        printf("Quitting\n");
    } else {
        number=atoi(p);
        printf("Number as int is %d\n", number);
    }
    free(p);

    return 0;
}

也就是说，通常更好的设计是将输入代码与解析输入的代码分开。 应该只以任何格式返回该行，并且您应该有一个单独的函数来检查这一点。readLine()allDigits()

同时检查“q”，您的代码将如下所示：

char *readLine(int nr) {
   int ch, i=0;
   char *string;
   string = malloc(nr);

   while((ch = getchar()) != '\n') {
       if(i<nr && isdigit(ch)!=0) {
          string[i++]=ch;
       }       
       else if (ch=='q' && i==0 && getchar()=='\n')
       {
        string[i++]=ch;
        return string;
       }                      
       else {
          string=NULL;
          return string;
       }
   }
   string[i]='\0';
   return string;
} 


int main(void) {
   int number;

   printf("Enter a number or q: ");
   char *p=readLine(100);

   if(p==NULL) {
      printf("Invalid input!\n");
   }
   else if (p[0]=='q' && p[1]=='\0')   
   {
      printf("You entered q\n");
   }
   
   else {
      number=atoi(p);
      printf("Number as int is %d ", number);
   }
   free(p);

   return 0;
}

在您的问题中，您表示您希望输入包含以下任一内容

• 仅数字，或
• 只有单个字符."q"

否则，输入无效，应被拒绝。

但是，在评论部分，您指出输入字符串也应该被拒绝。相反，您希望输入在某个最大值的范围内。"0"1

在这种情况下，使用该函数来确定输入是否有效是没有意义的，因为它会将该输入报告为有效，尽管您希望该输入被拒绝为无效。isdigit"0"

出于这个原因，我建议您首先使用函数 fgets 读取整行，然后检查该行的内容。如果该行由单个字符组成，则您知道输入有效。否则，您可以使用函数 strtol 尝试将输入转换为整数。q

如果此转换失败，那么您将知道输入无效。

但是，如果转换成功，则可以执行进一步的检查，以确定转换后的整数是否在所需范围内，例如，它是否在 的范围内。120

下面是一个示例：

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

#define MIN_INPUT_NUMBER 1
#define MAX_INPUT_NUMBER 20

void get_line_from_user( const char prompt[], char buffer[], int buffer_size );

int main( void )
{
    char line[200], *p;
    long num;

    //read exactly one line of input from the user
    get_line_from_user(
        "Please enter a number: ",
        line, sizeof line
    );

    //determine whether the user entered `"q"` as input
    if ( line[0] == 'q' && line[1] == '\0' )
    {
        printf( "Input is valid, user entered \"q\".\n" );
        exit( EXIT_SUCCESS );
    }

    //attempt to convert input to a number
    errno = 0;
    num = strtol( line, &p, 10 );

    //verify that at least one character was converted
    if ( p == line )
    {
        printf( "Error: Unable to convert input to integer!\n" );
        exit( EXIT_FAILURE );
    }

    //verify that input is not too small or too large to be
    //representable as a `long int`
    if ( errno == ERANGE )
    {
        printf( "Error: Input is outside the range of a long int!\n" );
        exit( EXIT_FAILURE );
    }

    //verify that either all characters were converted or that all
    //remaining characters are whitespace characters, so that input
    //such as "242q42" gets rejected
    for ( ; *p != '\0'; p++ )
    {
        if ( !isspace( (unsigned char)*p ) )
        {
            printf( "Error: Not all non-whitespace characters were converted!\n" );
            exit( EXIT_FAILURE );
        }
    }

    //verify that the converted integer is in the desired range
    if ( num < MIN_INPUT_NUMBER || num > MAX_INPUT_NUMBER )
    {
        printf(
            "Error: Input must be in the range %d to %d.\n",
            MIN_INPUT_NUMBER, MAX_INPUT_NUMBER
        );
        exit( EXIT_FAILURE );
    }

    //everything went ok, so print the result
    printf( "Input was valid! User entered %ld.\n", num );

    return EXIT_SUCCESS;
}

//This function will read exactly one line of input from the
//user. It will remove the newline character, if it exists. If
//the line is too long to fit in the buffer, then the function
//will automatically reprompt the user for input. On failure,
//the function will never return, but will print an error
//message and call "exit" instead.
void get_line_from_user( const char prompt[], char buffer[], int buffer_size )
{
    for (;;) //infinite loop, equivalent to while(1)
    {
        char *p;

        //prompt user for input
        fputs( prompt, stdout );

        //attempt to read one line of input
        if ( fgets( buffer, buffer_size, stdin ) == NULL )
        {
            printf( "Error reading from input!\n" );
            exit( EXIT_FAILURE );
        }

        //attempt to find newline character
        p = strchr( buffer, '\n' );

        //make sure that entire line was read in (i.e. that
        //the buffer was not too small to store the entire line)
        if ( p == NULL )
        {
            int c;

            //a missing newline character is ok if the next
            //character is a newline character or if we have
            //reached end-of-file (for example if the input is
            //being piped from a file or if the user enters
            //end-of-file in the terminal itself)
            if ( (c=getchar()) != '\n' && !feof(stdin) )
            {
                if ( c == EOF )
                {
                    printf( "Error reading from input!\n" );
                    exit( EXIT_FAILURE );
                }

                printf( "Input was too long to fit in buffer!\n" );

                //discard remainder of line
                do
                {
                    c = getchar();

                    if ( c == EOF )
                    {
                        //this error message will be printed if either
                        //a stream error or an unexpected end-of-file
                        //is encountered
                        printf( "Error reading from input!\n" );
                        exit( EXIT_FAILURE );
                    }

                } while ( c != '\n' );

                //reprompt user for input by restarting loop
                continue;
            }
        }
        else
        {
            //remove newline character by overwriting it with
            //null character
            *p = '\0';
        }

        //input was ok, so break out of loop
        break;
    }
}

此程序具有以下行为：

Please enter a number: 242q42
Error: Not all non-whitespace characters were converted!

Please enter a number: 5435q
Error: Not all non-whitespace characters were converted!

Please enter a number: qq
Error: Unable to convert input to integer!

Please enter a number: q
Input is valid, user entered "q".

Please enter a number: 50000000000000000000000000
Error: Input is outside the range of a long int!

Please enter a number: -23
Error: Input must be in the range 1 to 20.

Please enter a number: 0
Error: Input must be in the range 1 to 20.

Please enter a number: 22
Error: Input must be in the range 1 to 20.

Please enter a number: 17
Input was valid! User entered 17.

为了更改允许的范围，您可以更改线

#define MIN_INPUT_NUMBER 1
#define MAX_INPUT_NUMBER 20

在程序中。