Write a function called stringindexof, which accepts two strings: the first is a word and the second is a single character.

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The function should return the first index in the word at which the character exists or -1 if the character is not found. Do not use the built in String.indexOf() function.

Examples:

stringIndexOf('awesome', 'e') // 2
stringIndexOf('awesome', 'z') // -1

What I have tried:

JavaScript

function stringIndexOf (word, char) {
  for (let i=0; i < word.length; i++) {
    if (word[i] == char) {
      i+=1
      return char;
    }
  }
  return -1;
}

Posted 17-Oct-20 22:09pm

Member 14968336

Updated 31-Dec-22 10:11am

Patrice T

v3

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Patrice T · Answer 1 · 2020-10-17T22:53:00

Try something like:

JavaScript

function stringIndexOf (word, char) {
  for (let i=0; i < word.length; i++) {
    if (word[i] == char) {
      i+=1;
      return i;
    }
  }
  return -1;
}

Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

JavaScript Debugging[^]
Chrome DevTools | Web | Google Developers[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.

Member 15880909 · Answer 2 · 2022-12-31T10:11:00

Solution 3

for (let i=0; i < str.length; i++) {
if (str[i] == char) {
i+=0;
return i;
}
}
return -1;
}

Posted 31-Dec-22 10:11am

Member 15880909

Comments

Graeme_Grant 31-Dec-22 17:14pm

This was already answered and accepted back in October 2020. Please stick to current questions where help is needed.