Mixing
async/await nodejs funciontality
I have following code snippet with async/await.
async function test1 () {
setTimeout(() => {
console.log("1")
}, 2000);
}
async function test2 () {
setTimeout(() => {
console.log("2")
}, 1000);
}
async function test3 () {
setTimeout(() => {
console.log("3")
}, 1500);
}
async function test4 () {console.log("4")}
async function run () {
await test1()
await test2()
await test3()
await test4()
}
run()
When I explore above code snippet, I am expecting output as 1, 2, 3, 4. But I got 4, 2, 3, 1. Am I missed anything here?
Node Version v10.13.0
node.js async-await
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
add a comment |
I have following code snippet with async/await.
async function test1 () {
setTimeout(() => {
console.log("1")
}, 2000);
}
async function test2 () {
setTimeout(() => {
console.log("2")
}, 1000);
}
async function test3 () {
setTimeout(() => {
console.log("3")
}, 1500);
}
async function test4 () {console.log("4")}
async function run () {
await test1()
await test2()
await test3()
await test4()
}
run()
When I explore above code snippet, I am expecting output as 1, 2, 3, 4. But I got 4, 2, 3, 1. Am I missed anything here?
Node Version v10.13.0
node.js async-await
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
2
You've missed the fact that you aren't doing anything awaitable, in any of thoseasyncfunctions, so they'll execute in the order of their timeouts (which looks correct).
– James
Nov 22 '18 at 13:05
add a comment |
I have following code snippet with async/await.
async function test1 () {
setTimeout(() => {
console.log("1")
}, 2000);
}
async function test2 () {
setTimeout(() => {
console.log("2")
}, 1000);
}
async function test3 () {
setTimeout(() => {
console.log("3")
}, 1500);
}
async function test4 () {console.log("4")}
async function run () {
await test1()
await test2()
await test3()
await test4()
}
run()
When I explore above code snippet, I am expecting output as 1, 2, 3, 4. But I got 4, 2, 3, 1. Am I missed anything here?
Node Version v10.13.0
node.js async-await
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
I have following code snippet with async/await.
async function test1 () {
setTimeout(() => {
console.log("1")
}, 2000);
}
async function test2 () {
setTimeout(() => {
console.log("2")
}, 1000);
}
async function test3 () {
setTimeout(() => {
console.log("3")
}, 1500);
}
async function test4 () {console.log("4")}
async function run () {
await test1()
await test2()
await test3()
await test4()
}
run()
When I explore above code snippet, I am expecting output as 1, 2, 3, 4. But I got 4, 2, 3, 1. Am I missed anything here?
Node Version v10.13.0
node.js async-await
node.js async-await
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
asked Nov 22 '18 at 13:02
Syed Habib MSyed Habib M
1,2521426
1,2521426
2
You've missed the fact that you aren't doing anything awaitable, in any of thoseasyncfunctions, so they'll execute in the order of their timeouts (which looks correct).
– James
Nov 22 '18 at 13:05
add a comment |
2
You've missed the fact that you aren't doing anything awaitable, in any of thoseasyncfunctions, so they'll execute in the order of their timeouts (which looks correct).
– James
Nov 22 '18 at 13:05
2
2
You've missed the fact that you aren't doing anything awaitable, in any of those
async functions, so they'll execute in the order of their timeouts (which looks correct).– James
Nov 22 '18 at 13:05
You've missed the fact that you aren't doing anything awaitable, in any of those
async functions, so they'll execute in the order of their timeouts (which looks correct).– James
Nov 22 '18 at 13:05
add a comment |
1 Answer
1
active
oldest
votes
awaiting test1, etc. is the same as awaiting setTimeout(...) directly. setTimeout is not promise-based and isn't taken into account in promise chain.
await test1(), etc. result in one-tick delays, run() promise resolves instantly.
In order for the code to work as intended, it should be:
function test1 () {
return new Promise(resolve => setTimeout(() => {
console.log("1");
resolve();
}, 2000));
}
test1, etc. don't need to be async because they cannot benefit from a promise created by async function.
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
awaiting test1, etc. is the same as awaiting setTimeout(...) directly. setTimeout is not promise-based and isn't taken into account in promise chain.
await test1(), etc. result in one-tick delays, run() promise resolves instantly.
In order for the code to work as intended, it should be:
function test1 () {
return new Promise(resolve => setTimeout(() => {
console.log("1");
resolve();
}, 2000));
}
test1, etc. don't need to be async because they cannot benefit from a promise created by async function.
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
add a comment |
awaiting test1, etc. is the same as awaiting setTimeout(...) directly. setTimeout is not promise-based and isn't taken into account in promise chain.
await test1(), etc. result in one-tick delays, run() promise resolves instantly.
In order for the code to work as intended, it should be:
function test1 () {
return new Promise(resolve => setTimeout(() => {
console.log("1");
resolve();
}, 2000));
}
test1, etc. don't need to be async because they cannot benefit from a promise created by async function.
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
add a comment |
awaiting test1, etc. is the same as awaiting setTimeout(...) directly. setTimeout is not promise-based and isn't taken into account in promise chain.
await test1(), etc. result in one-tick delays, run() promise resolves instantly.
In order for the code to work as intended, it should be:
function test1 () {
return new Promise(resolve => setTimeout(() => {
console.log("1");
resolve();
}, 2000));
}
test1, etc. don't need to be async because they cannot benefit from a promise created by async function.
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
awaiting test1, etc. is the same as awaiting setTimeout(...) directly. setTimeout is not promise-based and isn't taken into account in promise chain.
await test1(), etc. result in one-tick delays, run() promise resolves instantly.
In order for the code to work as intended, it should be:
function test1 () {
return new Promise(resolve => setTimeout(() => {
console.log("1");
resolve();
}, 2000));
}
test1, etc. don't need to be async because they cannot benefit from a promise created by async function.
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
answered Nov 22 '18 at 13:09
estusestus
74.8k22109227
74.8k22109227
add a comment |
add a comment |
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2
You've missed the fact that you aren't doing anything awaitable, in any of those
asyncfunctions, so they'll execute in the order of their timeouts (which looks correct).– James
Nov 22 '18 at 13:05