Thread: 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

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

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

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]).

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