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javascript function arguments.length giving wrong value

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-1

concat.js

export default () => 
 const argsLength = arguments.length;
 if (argsLength === 0) 
 return [];
 

 const target = arguments[0];
 if (argsLength == 1) 
 return target;
 

 for (let i=1; i<argsLength; i++) 
 const src = arguments[i];
 for (let i=0; i<src.length; i++) 
 target.push(src[i]);
 
 

 return target;

test.js

describe('concat should copy', () => 
 it('src array into target array', () => 
 const result = concat(); // Not passgin any parameters.
 expect(result.length).to.equal(0);
 );
);

I am debugging a javascript function with no arguments. In the image, the arguments.length is clearly 0. But when it is initialized to a variable, it is becoming 5. Can someone explain what's going on here?

javascript

share|improve this question

edited Mar 9 at 3:24

lch

asked Mar 9 at 2:27

lchlch

83622250

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  How is it clearly zero? arguments comes from the number of parameters passed to the function not the number of parameters defined in the function. So, somewhere you are passing 5 parameters to this function.
  
  – Get Off My Lawn
  Mar 9 at 2:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @GetOffMyLawn I am not passing any arguments. Updated the post with code
  
  – lch
  Mar 9 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think you are seeing some unexpected behavior here because arrow functions do not have a local binding for arguments. changing to the regular function syntax should give you the correct 0 for the length
  
  – MimiEAM
  Mar 9 at 3:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MimiEAM Yes, its the arrow that is causing the issue. Switching to normal syntax solved it. Is there a way to achieve this using arrow functions?
  
  – lch
  Mar 9 at 3:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  stackoverflow.com/a/40924845/4127542 . this solved the problem
  
  – lch
  Mar 9 at 3:52
  

  
  
  
  

add a comment | 

-1

concat.js

export default () => 
 const argsLength = arguments.length;
 if (argsLength === 0) 
 return [];
 

 const target = arguments[0];
 if (argsLength == 1) 
 return target;
 

 for (let i=1; i<argsLength; i++) 
 const src = arguments[i];
 for (let i=0; i<src.length; i++) 
 target.push(src[i]);
 
 

 return target;

test.js

describe('concat should copy', () => 
 it('src array into target array', () => 
 const result = concat(); // Not passgin any parameters.
 expect(result.length).to.equal(0);
 );
);

I am debugging a javascript function with no arguments. In the image, the arguments.length is clearly 0. But when it is initialized to a variable, it is becoming 5. Can someone explain what's going on here?

javascript

share|improve this question

edited Mar 9 at 3:24

lch

asked Mar 9 at 2:27

lchlch

83622250

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  How is it clearly zero? arguments comes from the number of parameters passed to the function not the number of parameters defined in the function. So, somewhere you are passing 5 parameters to this function.
  
  – Get Off My Lawn
  Mar 9 at 2:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @GetOffMyLawn I am not passing any arguments. Updated the post with code
  
  – lch
  Mar 9 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think you are seeing some unexpected behavior here because arrow functions do not have a local binding for arguments. changing to the regular function syntax should give you the correct 0 for the length
  
  – MimiEAM
  Mar 9 at 3:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MimiEAM Yes, its the arrow that is causing the issue. Switching to normal syntax solved it. Is there a way to achieve this using arrow functions?
  
  – lch
  Mar 9 at 3:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  stackoverflow.com/a/40924845/4127542 . this solved the problem
  
  – lch
  Mar 9 at 3:52
  

  
  
  
  

add a comment | 

concat.js

export default () => 
 const argsLength = arguments.length;
 if (argsLength === 0) 
 return [];
 

 const target = arguments[0];
 if (argsLength == 1) 
 return target;
 

 for (let i=1; i<argsLength; i++) 
 const src = arguments[i];
 for (let i=0; i<src.length; i++) 
 target.push(src[i]);
 
 

 return target;

test.js

describe('concat should copy', () => 
 it('src array into target array', () => 
 const result = concat(); // Not passgin any parameters.
 expect(result.length).to.equal(0);
 );
);

I am debugging a javascript function with no arguments. In the image, the arguments.length is clearly 0. But when it is initialized to a variable, it is becoming 5. Can someone explain what's going on here?

javascript

share|improve this question

edited Mar 9 at 3:24

lch

asked Mar 9 at 2:27

lchlch

83622250

export default () => 
 const argsLength = arguments.length;
 if (argsLength === 0) 
 return [];
 

 const target = arguments[0];
 if (argsLength == 1) 
 return target;
 

 for (let i=1; i<argsLength; i++) 
 const src = arguments[i];
 for (let i=0; i<src.length; i++) 
 target.push(src[i]);
 
 

 return target;

describe('concat should copy', () => 
 it('src array into target array', () => 
 const result = concat(); // Not passgin any parameters.
 expect(result.length).to.equal(0);
 );
);

share|improve this question

edited Mar 9 at 3:24

lch

asked Mar 9 at 2:27

lchlch

83622250

share|improve this question

edited Mar 9 at 3:24

lch

asked Mar 9 at 2:27

lchlch

83622250

asked Mar 9 at 2:27

lchlch

83622250

asked Mar 9 at 2:27

lchlch

83622250