Update a value in std::map?

[desktop]

If I have:

std::map<int,int> m;

where I have:

m.insert(std::make_pair(1,2));

how do I update the value to 7 associated with key = 1?



I was thinking:

(*m.find(1)).second = 7;

But is there no simpler way?

 

[Ondra Holub]

If I have:

std::map<int,int> m;

where I have:

m.insert(std::make_pair(1,2));

how do I update the value to 7 associated with key = 1?

I was thinking:

(*m.find(1)).second = 7;

But is there no simpler way?

m[1] = 7;

 

[Ondra Holub]

If I have:

std::map<int,int> m;

where I have:

how do I update the value to 7 associated with key = 1?

I was thinking:

(*m.find(1)).second = 7;

But is there no simpler way?

m[1] = 7;

Please note, operator[] has "side effect". In case, that key value is
not present in map, new one is added. But for your usage it should be
ok.

 

[Chris ( Val )]

If I have:

std::map<int,int> m;

where I have:

m.insert(std::make_pair(1,2));

how do I update the value to 7 associated with key = 1?

I was thinking:

(*m.find(1)).second = 7;

But is there no simpler way?

Yes:

m[1]=7;

But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.

Use the map::find member to check it.

 

[desktop]

If I have:

std::map<int,int> m;

where I have:

m.insert(std::make_pair(1,2));

how do I update the value to 7 associated with key = 1?

I was thinking:

(*m.find(1)).second = 7;

But is there no simpler way?

Yes:

m[1]=7;

But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.

Use the map::find member to check it.

Ok if I just want to subtract 1 I can just do:

m[1] = m[1]-1;

 

[Chris ( Val )]

Chris ( Val ) wrote:

If I have:
std::map<int,int> m;
where I have:
m.insert(std::make_pair(1,2));
how do I update the value to 7 associated with key = 1?
I was thinking:
(*m.find(1)).second = 7;
But is there no simpler way?

Yes:

m[1]=7;

But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.

Use the map::find member to check it.

Ok if I just want to subtract 1 I can just do:

m[1] = m[1]-1;

m[ 1 ] -= 1;

 

[desktop]

Chris ( Val ) wrote:

If I have:
std::map<int,int> m;
where I have:
m.insert(std::make_pair(1,2));
how do I update the value to 7 associated with key = 1?
I was thinking:
(*m.find(1)).second = 7;
But is there no simpler way?
Yes:
m[1]=7;
But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.
Use the map::find member to check it.

Ok if I just want to subtract 1 I can just do:

m[1] = m[1]-1;

m[ 1 ] -= 1;

Ok but my suggestion also works right - currently not possible to check it.

 

[Guest]

Chris ( Val ) wrote:





If I have:
std::map<int,int> m;
where I have:
m.insert(std::make_pair(1,2));
how do I update the value to 7 associated with key = 1?
I was thinking:
(*m.find(1)).second = 7;
But is there no simpler way?
Yes:
m[1]=7;
But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.
Use the map::find member to check it.
--
Chris Val
Ok if I just want to subtract 1 I can just do:

m[1] = m[1]-1;

m[ 1 ] -= 1;

Ok but my suggestion also works right - currently not possible to check it.

Yes, or you can do just '--m[1];'.

 

[Mark P]

Chris ( Val ) wrote:





If I have:
std::map<int,int> m;
where I have:
m.insert(std::make_pair(1,2));
how do I update the value to 7 associated with key = 1?
I was thinking:
(*m.find(1)).second = 7;
But is there no simpler way?
Yes:
m[1]=7;
But check that you have a key with a value
of 1 before doing that, because if it does
not exist, it will be created.
Use the map::find member to check it.
--
Chris Val
Ok if I just want to subtract 1 I can just do:

m[1] = m[1]-1;

m[ 1 ] -= 1;

Ok but my suggestion also works right - currently not possible to check it.

It works, but it requires two lookups instead of one.

 

[Juha Nieminen]

It works, but it requires two lookups instead of one.

I'm wondering if any compiler is smart enough to optimize the second
lookup away.

 

[Guest]

I'm wondering if any compiler is smart enough to optimize the second
lookup away.

It would be hard, since there is not guarantee (from the language) that
the reference returned the second time will be the same as on the first
call. A human can quite easily see that it will be, but I am not sure if
the compiler can.