finding length of characters in an assigned char array ??

charlatanksk · 05-04-2008, 01:11 PM

hello , this is my first post here and i hope i can be helpful around here, so here is my question

lets say that i have a multidimensional array defined as char names[1] [1] , the first column is to write the name of the person and the second column is to write the surname. After i write the name, this is where the question begins actually, for every consonant i must replace it by 0 and for every vowel i must put 1 , for example, if i have a name such as Jack, it must be 0100 ,

i've tried to write something on my own , but it didn't work

my guess is that if i can find the number of the characters in the name entered, i can replace them by writing a for loop and a switch

iMalc · 05-04-2008, 01:25 PM

If you've got "char names[1] [1]" then you've already got a bug. That entire array holds as much as a single char, nothing more.
You want two names so the first number has to be at least 2.
You want names of more than zero characters long, so the second number has to be something sensible like 32.

charlatanksk · 05-04-2008, 01:29 PM

sorry for that mistake, i fixed that mistake , so
what needs to be done to find the length of the name that user entered?

vart · 05-04-2008, 01:45 PM

strlen function should help here

charlatanksk · 05-04-2008, 02:12 PM

ok , i wrote a program but it doesn't work,

i am sorry if there are terrible mistakes here, but i'm just a beginner

what i want the program to do is to have names[0] as aaa and to print out 222

Code:

#include <stdio.h>
#include <string.h>
#include <conio.h>
int main()
{
char names [1][20];
names[0]="aaa";
char amk=(char)names[0];
for(int i=0;i<=strlen(names[0]);i++){
switch(amk){
case 'a' :
printf("2");
break;
default:
printf("1");
break;
}
}
getch();
return 0;
}

it says Lvalue is required in function main

root4 · 05-04-2008, 02:29 PM

fixed version:

Code:

...
  char names [1][20]={{"aaa"}};
  for(int i=0;i<strlen(names[0]);i++){
    char amk=names[0][i];
    switch(amk){
      case 'a' :
      printf("2");
      break;
      default:
      printf("1");
      break;
    }
  }
...

you cannot initialize a char array with a string litteral after its allocation (names[0]="aaa")
do it during its allocation or by using strcpy() (for example) after its allocation.
Be careful with indexes [0..strlen()-1] instead of [0..strlen()].
At last, you must update the 'amk' value for each loop(char amk=names[0][i]).

charlatanksk · 05-04-2008, 02:34 PM

thank you for your help !!
i know it went a little out of topic, but i got my answer so thank you

05-04-2008, 01:11 PM	#1
charlatanksk Registered User Join Date: May 2008 Posts: 4	finding length of characters in an assigned char array ?? hello , this is my first post here and i hope i can be helpful around here, so here is my question lets say that i have a multidimensional array defined as char names[1] [1] , the first column is to write the name of the person and the second column is to write the surname. After i write the name, this is where the question begins actually, for every consonant i must replace it by 0 and for every vowel i must put 1 , for example, if i have a name such as Jack, it must be 0100 , i've tried to write something on my own , but it didn't work my guess is that if i can find the number of the characters in the name entered, i can replace them by writing a for loop and a switch
charlatanksk is offline

05-04-2008, 01:25 PM	#2
iMalc Algorithm Dissector Join Date: Dec 2005 Location: New Zealand Posts: 2,475	If you've got "char names[1] [1]" then you've already got a bug. That entire array holds as much as a single char, nothing more. You want two names so the first number has to be at least 2. You want names of more than zero characters long, so the second number has to be something sensible like 32. __________________ My homepage Advice: Take only as directed - If symptoms persist, please see your debugger
iMalc is offline

05-04-2008, 01:29 PM	#3
charlatanksk Registered User Join Date: May 2008 Posts: 4	sorry for that mistake, i fixed that mistake , so what needs to be done to find the length of the name that user entered?
charlatanksk is offline

05-04-2008, 01:45 PM	#4
vart CSharpener Join Date: Oct 2006 Posts: 5,242	strlen function should help here __________________ If I have eight hours for cutting wood, I spend six sharpening my axe.
vart is offline

05-04-2008, 02:12 PM	#5
charlatanksk Registered User Join Date: May 2008 Posts: 4	ok , i wrote a program but it doesn't work, i am sorry if there are terrible mistakes here, but i'm just a beginner what i want the program to do is to have names[0] as aaa and to print out 222 Code: #include <stdio.h> #include <string.h> #include <conio.h> int main() { char names [1][20]; names[0]="aaa"; char amk=(char)names[0]; for(int i=0;i<=strlen(names[0]);i++){ switch(amk){ case 'a' : printf("2"); break; default: printf("1"); break; } } getch(); return 0; } it says Lvalue is required in function main
charlatanksk is offline

05-04-2008, 02:29 PM	#6
root4 Registered User Join Date: Apr 2008 Posts: 278	fixed version: Code: ... char names [1][20]={{"aaa"}}; for(int i=0;i<strlen(names[0]);i++){ char amk=names[0][i]; switch(amk){ case 'a' : printf("2"); break; default: printf("1"); break; } } ... you cannot initialize a char array with a string litteral after its allocation (names[0]="aaa") do it during its allocation or by using strcpy() (for example) after its allocation. Be careful with indexes [0..strlen()-1] instead of [0..strlen()]. At last, you must update the 'amk' value for each loop(char amk=names[0][i]).
root4 is offline

05-04-2008, 02:34 PM	#7
charlatanksk Registered User Join Date: May 2008 Posts: 4	thank you for your help !! i know it went a little out of topic, but i got my answer so thank you
charlatanksk is offline

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