Why this generator function in JavaScript is stopping before any yield expression?

Code:

function* logGenerator() {

    console.log(0);

    console.log(1, yield);

    console.log(2, yield);

    console.log(3, yield);

}



var gen = logGenerator();

gen.next('alpha');

console.log('mark');

gen.next('beta');

Output:

0

mark

1 beta

but, why not:

0

1 alpha

mark

2 beta

In the line: gen.next('alpha');
the code only logs '0' and stops but does not reach the first yield expression, while the actual should have been '0n1 alpha'.

And 'alpha' is never logged.

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

thanx for help @Lloyd Nicholson

– PranshuKhandal
Jan 18 at 19:26

add a comment |

Code:

function* logGenerator() {

    console.log(0);

    console.log(1, yield);

    console.log(2, yield);

    console.log(3, yield);

}



var gen = logGenerator();

gen.next('alpha');

console.log('mark');

gen.next('beta');

Output:

0

mark

1 beta

but, why not:

0

1 alpha

mark

2 beta

In the line: gen.next('alpha');
the code only logs '0' and stops but does not reach the first yield expression, while the actual should have been '0n1 alpha'.

And 'alpha' is never logged.

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

thanx for help @Lloyd Nicholson

– PranshuKhandal
Jan 18 at 19:26

add a comment |

Code:

function* logGenerator() {

    console.log(0);

    console.log(1, yield);

    console.log(2, yield);

    console.log(3, yield);

}



var gen = logGenerator();

gen.next('alpha');

console.log('mark');

gen.next('beta');

Output:

0

mark

1 beta

but, why not:

0

1 alpha

mark

2 beta

In the line: gen.next('alpha');
the code only logs '0' and stops but does not reach the first yield expression, while the actual should have been '0n1 alpha'.

And 'alpha' is never logged.

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

Code:

function* logGenerator() {

    console.log(0);

    console.log(1, yield);

    console.log(2, yield);

    console.log(3, yield);

}



var gen = logGenerator();

gen.next('alpha');

console.log('mark');

gen.next('beta');

Output:

0

mark

1 beta

but, why not:

0

1 alpha

mark

2 beta

In the line: gen.next('alpha');
the code only logs '0' and stops but does not reach the first yield expression, while the actual should have been '0n1 alpha'.

And 'alpha' is never logged.

javascript

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

edited Jan 2 at 0:38

Lloyd Nicholson

296111

edited Jan 2 at 0:38

Lloyd Nicholson

296111

edited Jan 2 at 0:38

Lloyd Nicholson

296111

asked Jan 1 at 19:06

PranshuKhandal

430115

asked Jan 1 at 19:06

PranshuKhandal

430115

asked Jan 1 at 19:06

PranshuKhandal

430115

thanx for help @Lloyd Nicholson

– PranshuKhandal
Jan 18 at 19:26

add a comment |

thanx for help @Lloyd Nicholson

– PranshuKhandal
Jan 18 at 19:26

thanx for help @Lloyd Nicholson

– PranshuKhandal
Jan 18 at 19:26

add a comment |

1 Answer
1

active

oldest

votes

You're understanding of where the generator stops is not correct. To be fair, it's one of the most confusing aspects of generators. A generator stops immediately when it hits yield even if that's in the middle of an expression.

So in your example:

console.log(0);        // logs 0

console.log(1, yield); // evaluates the parameters to console.log, sees yield and stops before running console.log

You never make it to the console.log() because it hits yield while evaluating the parameters. It will continue and finish the log on the next next(). This is why generators are often "primed" with an initial next() call if they are going to depend on values passed in. The first call of next() can't pass a value.

Here's a simpler example showing the issue:

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

1

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

So in your example:

console.log(0);        // logs 0

console.log(1, yield); // evaluates the parameters to console.log, sees yield and stops before running console.log

Here's a simpler example showing the issue:

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

1

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

add a comment |

So in your example:

console.log(0);        // logs 0

console.log(1, yield); // evaluates the parameters to console.log, sees yield and stops before running console.log

Here's a simpler example showing the issue:

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

1

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

add a comment |

So in your example:

console.log(0);        // logs 0

console.log(1, yield); // evaluates the parameters to console.log, sees yield and stops before running console.log

Here's a simpler example showing the issue:

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

So in your example:

console.log(0);        // logs 0

console.log(1, yield); // evaluates the parameters to console.log, sees yield and stops before running console.log

Here's a simpler example showing the issue:

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

let test = "start"



function* logGenerator() {

  console.log("genertator starting")

  test = yield

}



var gen = logGenerator();

// call initial next. `val1` is lost to priming the generator

gen.next('val1');



// the generator stops before the assignment to test

// test is still "start"

console.log(test)



// only on the second call can you get a value to test

gen.next('val2');

console.log(test)

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

edited Jan 1 at 19:25

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

answered Jan 1 at 19:13

Mark Meyer

39.1k33162

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

1

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

add a comment |

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

1

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

thanks:-) sorry for so much late response! i'm new to stackoverflow

– PranshuKhandal
Jan 18 at 19:23

thanks for the help

– PranshuKhandal
Jan 18 at 19:30

add a comment |

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