Golang Function Call Without Parentheses
I am going through the Golang tutorials on their website and am confused by code similar to this that I've simplified and reproduced here:
package main
import (
"fmt"
"math"
)
func main()
a := math.Sqrt2
fmt.Println(a)
This prints 1.4142135623730951 in the sandbox. Replacing a := math.Sqrt2 with a := math.Sqrt(2) does the same thing but I'm confused how the function can be called without parentheses. math.Sqrt is not a function pointer here (there is no math.Sqrt2 function anyway, it's a function being passed without any parentheses. The function in the Go documentation here is listed as: func Sqrt(x float64) float64 i.e. with the parameter. So how does that work? Is it just because math.Sqrt() is a simplistic function that Go can assume it's a float64 without the parentheses passed? Am I missing something?
If it helps, I found this phenomenon here in the tutorials on line 14, originally. If anyone could explain this feature to me, that would be awesome. I'd love to learn about it.
function go
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
add a comment |
I am going through the Golang tutorials on their website and am confused by code similar to this that I've simplified and reproduced here:
package main
import (
"fmt"
"math"
)
func main()
a := math.Sqrt2
fmt.Println(a)
This prints 1.4142135623730951 in the sandbox. Replacing a := math.Sqrt2 with a := math.Sqrt(2) does the same thing but I'm confused how the function can be called without parentheses. math.Sqrt is not a function pointer here (there is no math.Sqrt2 function anyway, it's a function being passed without any parentheses. The function in the Go documentation here is listed as: func Sqrt(x float64) float64 i.e. with the parameter. So how does that work? Is it just because math.Sqrt() is a simplistic function that Go can assume it's a float64 without the parentheses passed? Am I missing something?
If it helps, I found this phenomenon here in the tutorials on line 14, originally. If anyone could explain this feature to me, that would be awesome. I'd love to learn about it.
function go
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
10
math.Sqrt2is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go
– zerkms
Nov 14 '18 at 22:15
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25
add a comment |
I am going through the Golang tutorials on their website and am confused by code similar to this that I've simplified and reproduced here:
package main
import (
"fmt"
"math"
)
func main()
a := math.Sqrt2
fmt.Println(a)
This prints 1.4142135623730951 in the sandbox. Replacing a := math.Sqrt2 with a := math.Sqrt(2) does the same thing but I'm confused how the function can be called without parentheses. math.Sqrt is not a function pointer here (there is no math.Sqrt2 function anyway, it's a function being passed without any parentheses. The function in the Go documentation here is listed as: func Sqrt(x float64) float64 i.e. with the parameter. So how does that work? Is it just because math.Sqrt() is a simplistic function that Go can assume it's a float64 without the parentheses passed? Am I missing something?
If it helps, I found this phenomenon here in the tutorials on line 14, originally. If anyone could explain this feature to me, that would be awesome. I'd love to learn about it.
function go
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
I am going through the Golang tutorials on their website and am confused by code similar to this that I've simplified and reproduced here:
package main
import (
"fmt"
"math"
)
func main()
a := math.Sqrt2
fmt.Println(a)
This prints 1.4142135623730951 in the sandbox. Replacing a := math.Sqrt2 with a := math.Sqrt(2) does the same thing but I'm confused how the function can be called without parentheses. math.Sqrt is not a function pointer here (there is no math.Sqrt2 function anyway, it's a function being passed without any parentheses. The function in the Go documentation here is listed as: func Sqrt(x float64) float64 i.e. with the parameter. So how does that work? Is it just because math.Sqrt() is a simplistic function that Go can assume it's a float64 without the parentheses passed? Am I missing something?
If it helps, I found this phenomenon here in the tutorials on line 14, originally. If anyone could explain this feature to me, that would be awesome. I'd love to learn about it.
function go
function go
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
asked Nov 14 '18 at 22:14
HarpagornisHarpagornis
193
193
10
math.Sqrt2is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go
– zerkms
Nov 14 '18 at 22:15
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25
add a comment |
10
math.Sqrt2is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go
– zerkms
Nov 14 '18 at 22:15
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25
10
10
math.Sqrt2 is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go– zerkms
Nov 14 '18 at 22:15
math.Sqrt2 is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go– zerkms
Nov 14 '18 at 22:15
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
There is nothing special happening here. math.Sqrt2 is a constant. You can find the other constants in the math package in the docs.
In general, go doesn't really have any "magic". So if something feels a bit magical, its more than likely just a misunderstanding.
answered Nov 14 '18 at 22:36
poypoy
6,50763364
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part ofreturn(multiple value returns). To me - that's totally counter-intuitive magic
– zerkms
Nov 15 '18 at 20:06
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is nothing special happening here. math.Sqrt2 is a constant. You can find the other constants in the math package in the docs.
In general, go doesn't really have any "magic". So if something feels a bit magical, its more than likely just a misunderstanding.
answered Nov 14 '18 at 22:36
poypoy
6,50763364
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part ofreturn(multiple value returns). To me - that's totally counter-intuitive magic
– zerkms
Nov 15 '18 at 20:06
add a comment |
There is nothing special happening here. math.Sqrt2 is a constant. You can find the other constants in the math package in the docs.
In general, go doesn't really have any "magic". So if something feels a bit magical, its more than likely just a misunderstanding.
answered Nov 14 '18 at 22:36
poypoy
6,50763364
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part ofreturn(multiple value returns). To me - that's totally counter-intuitive magic
– zerkms
Nov 15 '18 at 20:06
add a comment |
There is nothing special happening here. math.Sqrt2 is a constant. You can find the other constants in the math package in the docs.
In general, go doesn't really have any "magic". So if something feels a bit magical, its more than likely just a misunderstanding.
answered Nov 14 '18 at 22:36
poypoy
6,50763364
There is nothing special happening here. math.Sqrt2 is a constant. You can find the other constants in the math package in the docs.
In general, go doesn't really have any "magic". So if something feels a bit magical, its more than likely just a misunderstanding.
answered Nov 14 '18 at 22:36
poypoy
6,50763364
answered Nov 14 '18 at 22:36
poypoy
6,50763364
answered Nov 14 '18 at 22:36
poypoy
6,50763364
answered Nov 14 '18 at 22:36
poypoy
6,50763364
6,50763364
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part ofreturn(multiple value returns). To me - that's totally counter-intuitive magic
– zerkms
Nov 15 '18 at 20:06
add a comment |
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part ofreturn(multiple value returns). To me - that's totally counter-intuitive magic
– zerkms
Nov 15 '18 at 20:06
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
Got it! I was overthinking it! Thanks a lot!
– Harpagornis
Nov 15 '18 at 19:26
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part of
return (multiple value returns). To me - that's totally counter-intuitive magic– zerkms
Nov 15 '18 at 20:06
"go doesn't really have any "magic"" --- unless it does: expressions may behave differently depending on the context. Eg: as a part of a composite expression vs as a part of
return (multiple value returns). To me - that's totally counter-intuitive magic– zerkms
Nov 15 '18 at 20:06
add a comment |
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10
math.Sqrt2is a constant golang.org/pkg/math/#pkg-constants golang.org/src/math/const.go– zerkms
Nov 14 '18 at 22:15
Ahh, I see. Thanks!
– Harpagornis
Nov 15 '18 at 19:25