如何使用 C 中的基本命令将数字转换为单词

Question

嗨，我真的是 C 的新手。
我正在做一个练习，我必须编写一个程序，该程序接受两个输入整数（包括 -10 和 10 之间的整数）并输出总和。
但是，显示的从 -10 到 10 的所有数字都将显示为它们各自的单词，任何其他数字都将是数字数字。
我使用的唯一命令是 if's、else if's、else、printf、scanf、int。

一个例子是：

Enter your two integers: 9 2
nine + two = eleven

Enter your two integers: -9 4
negative nine + four = negative five

Enter your two integers: -10 -2
negative ten + negative two = -12

感谢任何帮助。这是我到目前为止所做的，但我不确定，因为它看起来很乏味，但我的朋友说她做的相当快。

#include <stdio.h>

int main(void){
    int number1;
    int number2;
    int sum;
    printf("Please enter two integers: ");
    scanf("%d",&number1,&number2);
    sum = number1 + number2
    if (number1 > 0 && number2 > 0) {
        if (number1 == 1 && number2 == 1) {
            printf("one + one = two");
        }
    }
}

Answer 1

我使用的唯一命令是 if's、else if's、else、printf、scanf、int。

我同意你的观点，这看起来很乏味。

我想说的一点是，您可以单独计算每个数字 - 这将减少乏味。

例如，修改您的代码...（请注意，这不是可编译的 - 只是稍作修改以提供建议）

#include <stdio.h>

int main(void){
    int number1;
    int number2;
    int sum;
    int inputs;

    printf("Please enter two integers: ");

     // scanf replaced with new one including while
     // the first statement below will work provided the user provides the 
     // correct input data....
     //
     // scanf("%d %d",&number1,&number2); // added second %d - need two %d  one for each number
    //
    // would be better to this next scanf command below...
    // if scanf doesn't work due to bad input it will tell the user
    // while (scanf("%d %d",&number1,&number2) !=2) printf("incorrect input\nPlease enter two integers: "); 
    // but this uses 'while' which you haven't used yet if I understand correctly
    // 
    inputs = scanf("%d %d",&number1,&number2);
    if (inputs!=2) {
      printf("Incorrect input data\n"); 
      return 1;
    }
    // if scanf works inputs will have the value 2
    //
    // this if statement stops the program if inputs is not equal to 2
    // 
    sum = number1 + number2

    // the next bit suggested by Groo handles negative numbers nicely....
    if (number1 < 0) { 
      number1 = -number; printf("negative "); 
    }
    // if (number1 > 0 && number2 > 0) 
    if (number1 == 0) {
       printf("zero");
    }
    else if (number1 == 1) {
        printf("one");
    } else if (number1 == 2) {
      printf("two");
    } else if(number1 == 3) {
      printf("three");
    } else if(number1 == 4) {
      printf("four");
    } else if(number1 == 5) {
      printf("five");
    } else if(number1 == 6) {
      printf("six");
    } else if (number1 == 7) {
      printf("seven");
    } else if(number1 == 8) {
      printf("eight");
    } else if(number1 == 9) {
      printf("nine");
    } else if (number1 == 10) {
      printf("ten");
    } else  {
      printf("unknown number");
    } 
    printf(" + ");
     // now insert similar if statements to print number2 ....
      ....

  return 0;

}

希望这是一个有用的建议 - 如果我必须这样做，我会使用数组，我想你很快就会了解。

Answer 2

这里的实际算法是简单的部分，只需声明一个指向字符串字面量的指针数组并将其用作查找表：

const char* TEXTUAL [11] = 
{
  "zero",
  "one",
  "two",
  "three",
  "four",
  "five",
  "six",
  "seven",
  "eight",
  "nine",
  "ten"
};  

printf("%s", TEXTUAL[value]);

这里最难的部分是所有的错误检查。您必须从用户那里读取两个整数，检查它们是否有效并且在 -10 到 10 的范围内。然后检查结果是否也在范围内。

我做了一个例子，你可以看看提示。如何在 C 中安全地读取整数是它自己的一章，我创建了一个函数read_2_int，它以安全的方式读取用户输入。如果您是初学者，请暂时忽略该功能并使用scanf("%d", &var)。

为了处理负数，我使用函数labs 来获取整数的绝对值。

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

static void halt_and_catch_fire (const char* err);
static void read_2_int (int* i1, int* i2);

int main (void)
{
  int n1;
  int n2;

  printf("Enter your two integers: ");
  read_2_int(&n1, &n2);


  if(n1 > 10 || n1 < -10 || n2 > 10 || n2 < -10)
  {
    halt_and_catch_fire("Only numbers from -10 to 10 supported.");
  }
  if(n1+n2 > 10 || n1+n2 < -10)
  {
    halt_and_catch_fire("The sum must be from -10 to 10.");
  }


  const char* TEXTUAL [11] = 
  {
    "zero",
    "one",
    "two",
    "three",
    "four",
    "five",
    "six",
    "seven",
    "eight",
    "nine",
    "ten"
  };  


  if(n1 < 0){ printf("minus "); }
  printf("%s + ", TEXTUAL[labs(n1)] ); // labs = absolute value
  if(n2 < 0){ printf("minus "); }
  printf("%s = ", TEXTUAL[labs(n2)] );
  if(n1 + n2 < 0) { printf("minus "); }
  printf("%s\n", TEXTUAL[labs(n1+n2)] );
}


static void halt_and_catch_fire (const char* err)
{
  /* some manner of error handling here */
  fprintf(stderr, err);
  exit(EXIT_FAILURE);
}

static void read_2_int (int* i1, int* i2)
{
  char input[10];

  if(fgets(input, 10, stdin) == NULL)
  {
    halt_and_catch_fire ("You have to type two integers.");
  }

  char* endptr;
  *i1 = strtol(input, &endptr, 10);
  if(endptr == input)
  {
    halt_and_catch_fire ("Bad number format");
  }

  char* next = endptr;
  *i2 = strtol(next, &endptr, 10);
  if(endptr == next)
  {
    halt_and_catch_fire ("You have to type two integers.");
  }
}

Answer 3

我希望以下代码有用：

#include <stdio.h>

int main(void){
    int number1;
    int number2;
    int sum;
    char *digits[] = { "zero", "one", "two", "three", "four",
                       "five", "six", "seven", "eight", "nine", "ten"};

    printf("Please enter two integers: ");
    scanf("%d",&number1);
    scanf("%d",&number2);
    sum = number1 + number2;

    if(number1<0){
        printf("negative ");
        number1=-number1;
    }
    printf("%s",digits[number1]);
    printf(" + ");

    if(number2<0){
        printf("negative ");
        number2=-number2;
    }
    printf("%s",digits[number2]);
    printf(" = ");
    printf("%d",sum);
}

Answer 4

我编写了一个适用于从 0 到 99,999 的所有数字的程序。 代码如下：

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

int main() {
int n,i;
int rem,r2,k=0;
char *word[100000];
char *word_temp[100000];
char *units[]={" zero"," one"," two"," three"," four"," five"," six"," seven"," eight"," nine"};
char *teens[]={" ten"," eleven"," twelve"," thirteen"," fourteen"," fifteen"," sixteen"," seventeen"," eighteen"," nineteen"};
char *tenths[]={" twenty"," thirty"," forty"," fifty"," sixty"," seventy"," eighty"," ninety"};
char *hundredth[]={" hundred"};
char *thousandth[]={" thousand"};
printf("\nenter the number(only positive and less than one lakh): ");
scanf("%d",&n);

if(n==0){
    word[k++]=units[0];

    printf("\nthe number in words: ");
    for(i=(k-1);i>=0;i--)
        printf("%s",word[i]);
}

else if(n>0&&n<=99999){
    while(n>0) {
        word_temp[k]="";

        if((n>999)&&(n<=99999)){
            rem=n/1000;

            if((rem>19)&&(rem<=99)){
                if((rem%10)!=0){
                    word_temp[++k]=units[rem%10];
                    word[k]=word_temp[k];
                    k--;
                }
                word_temp[k]=tenths[(rem/10)-2];
                word[k]=word_temp[k];
                if((rem%10)!=0)
                    k+=2;
                else
                    k++;
            }

            else {
                if(rem<10)
                    word_temp[k]=units[rem];

                else if(rem>9&&rem<20)
                    word_temp[k]=teens[rem%10];  
                    word[k]=word_temp[k];
                    k++;
            }
            word_temp[k]=thousandth[0];
            word[k]=word_temp[k];
            k++;
            n=n-(rem*1000);
        }

        else if((n>99)&&(n<=999)){
            rem=n/100;
            word_temp[k]=units[rem];
            word[k]=word_temp[k];
            k++;
            word_temp[k]=hundredth[0];
            word[k]=word_temp[k];
            k++;
            word[k++]=" and";
            n=n-(rem*100);
        }

        else if((n>19)&&(n<=99)){
            rem=n/10;
            word_temp[k]=tenths[rem-2];
            word[k]=word_temp[k];
            k++;
            n=n-(rem*10);
        }

        else if((n>9)&&(n<=19)){
            rem=n;
            word_temp[k]=teens[rem%10];
            word[k]=word_temp[k];
            k++;
            n=n-(rem*1);
            break;
        }

        else if((n>0)&&(n<=9)){
            rem=n%10;
            word_temp[k]=units[rem];
            word[k]=word_temp[k];
            k++;
            n=0;
            break;
        }

    }

    printf("\nthe number in words:\n ");
    for(i=0;i<=k-1;i++)
        printf("%s ",word[i]);
    }
    else {
         printf("\nvalue out of range...Exiting...");
         exit(0);
    }

    printf("\n\nprocess terminated... ");
    }

Answer 5

我建议做一个辅助功能

void numberName(int number){

  if (number < 0){
      printf("negative ");
      number=number*(-1);
  }
  if(number == 0){
      printf("zero");
  }else if (number == 1){
      printf("one");
  }else if (number == 2){
      printf("two");
  }else if (number == 3){
      printf("three");
  }else if (number == 4){
      printf("four");
  }else if (number == 5){
      printf("five");
  }else if (number == 6){
      printf("six");
  }else if (number == 7){
      printf("seven");
  }else if (number == 8){
      printf("eight");
  }else if (number == 9){
      printf("nine");
  }else if (number == 10){
      printf("ten");
  }
}

并在您的 main() 代码中像这样调用此函数：

numberName(number1);
printf(" + ");
numberName(number2);
printf(" = ");
numberName(sum);
printf("\n");

顺便看看你的代码，扫描功能应该是：

scanf("%d%d",&number1,&number2);