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program that will input a number and display it in words

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program that will input a number and display it in words

 
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  #1
Dec 12th, 2007
im a 1st year comsci student and we are given an assignment that will display in words the number that is entered using logical operators and what we've learned so far(printf, scanf,etc.)
I only have problems in displaying the numbers from 11-19 (when I input 12 is displays two twelve), and when i got through it comes another problem with the ones digits (when I input 1 nothing comes out).
pls help..

here's the code so far..
C Syntax (Toggle Plain Text) 
  1. #include<stdio.h>
  2. #include<conio.h>
  3. main()
  4. {
  5. int num, a,w,x,y,z;
  6. clrscr();
  7. printf("Enter a number: ");
  8. scanf("%d", &num);
  9.  
  10. a=num/1000;
  11. if (a==1)
  12.    printf("one thousand");
  13. else if (a==2)
  14.    printf("two thousand");
  15. else if (a==3)
  16.    printf("three thousand");
  17. else if (a==4)
  18.    printf("four thousand");
  19. else if (a==5)
  20.    printf("five thousand");
  21. else if (a==6)
  22.    printf("six thousand");
  23. else if (a==7)
  24.    printf("seven thousand");
  25. else if (a==8)
  26.    printf("eight thousand");
  27. else if (a==9)
  28.    printf("nine thousand");
  29.  
  30. y=num%1000;
  31. if (y>=100&&y<200)
  32.    printf(" one hundred");
  33. else if (y>=200&&y<300)
  34.    printf(" two hundred");
  35. else if (y>=300&&y<400)
  36.    printf(" three hundred");
  37. else if (y>=400&&y<500)
  38.    printf(" four hundred");
  39. else if (y>=500&&y<600)
  40.    printf(" five hundred");
  41. else if (y>=600&&y<700)
  42.    printf(" six hundred");
  43. else if (y>=700&&y<800)
  44.    printf(" seven hundred");
  45. else if (y>=800&&y<900)
  46.    printf(" eight hundred");
  47. else if (y>=900&&y<1000)
  48.    printf(" nine hundred");
  49.  
  50. z=num%100;
  51. if (z>=20&&z<30)
  52.    printf( " twenty");
  53. else if (z>=30&&z<40)
  54.    printf( " thirty");
  55. else if (z>=40&&z<50)
  56.    printf( " forty");
  57. else if (z>=50&&z<60)
  58.    printf( " fifty");
  59. else if (z>=60&&z<70)
  60.    printf( " sixty");
  61. else if (z>=70&&z<80)
  62.    printf( " seventy");
  63. else if (z>=80&&z<90)
  64.    printf( " eighty");
  65. else if (z>=90&&z<100)
  66.    printf( " ninety");
  67.  
  68. if (x<10&&x>20)
  69.  
  70. x=z%10;
  71. if (x==1)
  72.    printf(" one");
  73. else if (x==2)
  74.    printf(" two");
  75. else if (x==3)
  76.    printf(" three");
  77. else if (x==4)
  78.    printf(" four");
  79. else if (x==5)
  80.    printf(" five");
  81. else if (x==6)
  82.    printf(" six");
  83. else if (x==7)
  84.    printf(" seven");
  85. else if (x==8)
  86.    printf(" eight");
  87. else if (x==9)
  88.    printf(" nine");
  89.  
  90. if (x>10&&x<20)
  91.   { printf("");}
  92.  
  93. w=num;
  94. if (w==10)
  95.    printf("ten");
  96. else if (w==11)
  97.    printf(" eleven");
  98. else if (w==12)
  99.    printf(" twelve");
  100. else if (w==13)
  101.    printf(" thirteen");
  102. else if (w==14)
  103.    printf(" fourteen");
  104. else if (w==15)
  105.    printf(" fifteen");
  106. else if (w==16)
  107.    printf(" sixteen");
  108. else if (w==17)
  109.    printf(" seventeen");
  110. else if (w==18)
  111.    printf(" eighteen");
  112. else if (w==19)
  113.    printf(" nineteen");
  114.  
  115. getch();
  116. }
  117. 
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Re: program that will input a number and display it in words

 
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  #2
Dec 12th, 2007
Look carefully at this section of your code:
C Syntax (Toggle Plain Text) 
  1. if (x<10&&x>20)
  2.  
  3. x=z%10;
  4. if (x==1)
  5.    printf(" one");
  6. ...

You also might want to consider using more spaces in each line to make the code more readable. Something like:

C Syntax (Toggle Plain Text) 
  1. z = num % 100;
  2. if (z >= 20  &&  z < 30)
  3. ...
  4. else if (z >= 40  &&  z < 50)
  5.    printf( " forty");
  6. else if (z >= 50  &&  z < 60)
  7.    printf( " fifty");
  8. else if (z >= 60  &&  z < 70)
  9.    printf( " sixty");
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Re: program that will input a number and display it in words

 
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  #3
Dec 12th, 2007
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Originally Posted by WaltP View Post
Look carefully at this section of your code:
C Syntax (Toggle Plain Text) 
  1. if (x<10&&x>20)
  2.  
  3. x=z%10;
  4. if (x==1)
  5.    printf(" one");
  6. ...
at first I didn't have the
C Syntax (Toggle Plain Text) 
  1. if (x<10&&x>20)
thats the time where the output for 12 is two twelve and 512 is five hundred two.. but when its there the ones digit are not displayed 1 is blank and 554 is five hundred fifty..
its like every time I fix one problem , another one comes out..
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Re: program that will input a number and display it in words

 
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  #4
Dec 12th, 2007
What WaltP is trying to say is that: if (x<10&&x>20) is wrong:
This statement means: if x is less then 10 AND bigger then 20. What number do you know that is smaller then 10 and bigger then 20?
Just replace it with : if (x<10)
And another thing: what happens when your number ends with a zero?

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Originally Posted by makubexiii View Post
its like every time I fix one problem , another one comes out..
Welcome to the wonderfull world of programming

Niek
Last edited by niek_e; Dec 12th, 2007 at 8:06 am.
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Re: program that will input a number and display it in words

 
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  #5
Dec 12th, 2007
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Originally Posted by niek_e View Post
What WaltP is trying to say is that: if (x<10&&x>20) is wrong:
This statement means: if x is less then 10 AND bigger then 20. What number do you know that is smaller then 10 and bigger then 20?
Just replace it with : if (x<10)
And another thing: what happens when your number ends with a zero?


Welcome to the wonderfull world of programming

Niek
the numbers that ends in zero are fine.. 120 will just display one hundred twenty..
i think I need a code that if its between 11-19 it will display the phrase and not display the ones digit. but if its outside 11-19 it will disregard the 11-19 codes..
if i use if (x<10) it still wont display the ones digits

the world of programming sure is wonderful.. it really cracks my head..
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Re: program that will input a number and display it in words

 
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  #6
Dec 12th, 2007
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C Syntax (Toggle Plain Text) 
  1. if (x<10&&x>20)
  2.  
  3. x=z%10;
I tried to run your code and came across the above. You're checking if x<10 BEFORE it is assigned a value... This will crash your program.
Change it to this:
c Syntax (Toggle Plain Text) 
  1. x=z%10;
  2. if (x<10)
And you're done! (well..almost anyway. Try 12 for example. This is all due to placement of if's)

clrscr() is not a standard C function not is getch(). I guess you're using the ancient turbo c compiler? Anyway, removing those lines (and conio.h) will make you're code standard C so it will compile on other compilers. Replace getch() with getchar();

regards Niek
Last edited by niek_e; Dec 12th, 2007 at 11:46 am.
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Re: program that will input a number and display it in words

 
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  #7
Dec 12th, 2007
You might also consider using arrays for your 'words'. It would make your code shorter. For example:
c Syntax (Toggle Plain Text) 
  1. char *number[]= { "", "one", "two", "three", "four"... "nine"};
  2. ...
  3. x = z % 10;
  4. printf( " %s", number[x]);
You can do this for all your numbers, including the teens and 10-90.
The 'zero' is blank because you never really output 0 in word form.
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2) Tomorrow never comes
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Re: program that will input a number and display it in words

 
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Dec 12th, 2007
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Originally Posted by WaltP View Post
You might also consider using arrays for your 'words'. It would make your code shorter. For example:
c Syntax (Toggle Plain Text) 
  1. char *number[]= { "", "one", "two", "three", "four"... "nine"};
  2. ...
  3. x = z % 10;
  4. printf( " %s", number[x]);
You can do this for all your numbers, including the teens and 10-90.
The 'zero' is blank because you never really output 0 in word form.
the only problem with this is that it hasn't been taught to us yet (the *number[] part)..
so we have to use only whatever has been taught to us
i tried using nested if's but all those {} are making me dizzy, i keep getting errors and i cant keep track of my variables..

@niek_e
yes im using the ancient turbo c. thats what we'rt told to use..
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Re: program that will input a number and display it in words

 
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  #9
Dec 13th, 2007
just got back from school..
i got the 10-19 part ok now.. before when i type 312 it says three hundred two but its fixed now..
the only problem left is that it wont display ones digits..
i've gone over reviewing this many times but i cant figure out whats wrong..
heres the code now

C Syntax (Toggle Plain Text) 
  1. #include<stdio.h>
  2. #include<conio.h>
  3. main()
  4. {
  5. int num, a,w,x,y,z;
  6. clrscr();
  7. printf("Enter a number: ");
  8. scanf("%d", &num);
  9.  
  10.  
  11. a=num/1000;
  12. if (a==1)
  13.    printf("one thousand");
  14. else if (a == 2)
  15.    printf("two thousand");
  16. else if (a== 3)
  17.    printf("three thousand");
  18. else if (a == 4)
  19.    printf("four thousand");
  20. else if (a == 5)
  21.    printf("five thousand");
  22. else if (a==6)
  23.    printf("six thousand");
  24. else if (a == 7)
  25.    printf("seven thousand");
  26. else if (a == 8)
  27.    printf("eight thousand");
  28. else if (a == 9)
  29.    printf("nine thousand");
  30.  
  31. y=num % 1000;
  32. if (y >= 100 && y < 200)
  33.    printf(" one hundred");
  34. else if (y >= 200 && y < 300)
  35.    printf(" two hundred");
  36. else if (y >= 300 && y < 400)
  37.    printf(" three hundred");
  38. else if (y >= 400 && y < 500)
  39.    printf(" four hundred");
  40. else if (y >= 500 && y < 600)
  41.    printf(" five hundred");
  42. else if (y >= 600 && y < 700)
  43.    printf(" six hundred");
  44. else if (y >= 700 && y < 800)
  45.    printf(" seven hundred");
  46. else if (y >= 800 && y < 900)
  47.    printf(" eight hundred");
  48. else if (y >= 900 && y < 1000)
  49.    printf(" nine hundred");
  50.  
  51. z=num%100;
  52. if (z >= 20 && z < 30)
  53.    printf( " twenty");
  54. else if (z >= 30 && z < 40)
  55.    printf( " thirty");
  56. else if (z >= 40 && z < 50)
  57.    printf( " forty");
  58. else if (z >= 50 && z < 60)
  59.    printf( " fifty");
  60. else if (z >= 60 && z < 70)
  61.    printf( " sixty");
  62. else if (z >= 70 && z < 80)
  63.    printf( " seventy");
  64. else if (z >= 80 && z < 90)
  65.    printf( " eighty");
  66. else if (z >= 90 && z < 100)
  67.    printf( " ninety");
  68. else if (z >= 10 && z < 20)
  69.   {w = z;}
  70.  
  71. if (w == 10)
  72.    printf(" ten");
  73. else if (w == 11)
  74.    printf(" eleven");
  75. else if (w ==12)
  76.    printf(" twelve");
  77. else if (w == 13)
  78.    printf(" thirteen");
  79. else if (w == 14)
  80.    printf(" fourteen");
  81. else if (w == 15)
  82.    printf(" fifteen");
  83. else if (w == 16)
  84.    printf(" sixteen");
  85. else if (w == 17)
  86.    printf(" seventeen");
  87. else if (w == 18)
  88.    printf(" eighteen");
  89. else if (w == 19)
  90.    printf(" nineteen");
  91.  
  92.  
  93. if (z < 10 || z >= 20)
  94.    {x = z % 10;}
  95. else if (x == 1)
  96.    printf(" one");
  97. else if (x == 2)
  98.    printf(" two");
  99. else if (x == 3)
  100.    printf(" three");
  101. else if (x == 4)
  102.    printf(" four");
  103. else if (x == 5)
  104.    printf(" five");
  105. else if (x == 6)
  106.    printf(" six");
  107. else if (x == 7)
  108.    printf(" seven");
  109. else if (x == 8)
  110.    printf(" eight");
  111. else if (x == 9)
  112.    printf(" nine");
  113.  
  114. getch();
  115. }
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Re: program that will input a number and display it in words

 
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  #10
Dec 13th, 2007
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Originally Posted by makubexiii View Post
the only problem with this is that it hasn't been taught to us yet (the *number[] part)..
so we have to use only whatever has been taught to us
i tried using nested if's but all those {} are making me dizzy, i keep getting errors and i cant keep track of my variables..
Then you're probably not formatting them correctly...

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Originally Posted by makubexiii View Post
just got back from school..
So?

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Originally Posted by makubexiii View Post
i got the 10-19 part ok now.. before when i type 312 it says three hundred two but its fixed now..
the only problem left is that it wont display ones digits..
i've gone over reviewing this many times but i cant figure out whats wrong..
Again, look carefully at this piece of code -- before niek_e gets here and tells you what I mean
C Syntax (Toggle Plain Text) 
  1. if (z < 10 || z >= 20)
  2.    {x = z % 10;}
  3. else if (x == 1)
  4.    printf(" one");
  5. else if (x == 2)
  6.    printf(" two");
  7. ...
How does this code allow you to get to printf(" one") ?
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