Stroustrup section 5.5 "vector of struct"

A

arnuld

this is the code:

-------------------------------------------------------------------------

#include <iostream>
#include <string>
#include <vector>

struct Pair {
std::string name;
double val;
};

std::vector<Pair> pairs;

double& value(const std::string s) {
for(int i=0; i < pairs.size(); i++)
if(s == pairs.name) return pairs.val;

Pair p = {s, -1.1};
pairs.push_back(p); // pair added at the end

return pairs[pairs.size() - 1].val;
}


int main() {

Pair p0 = {"p0", 0.0};
Pair p1 = {"p1", 1.0};
Pair p2 = {"p2", 2.0};

// add to "pairs"
pairs.push_back(p0);
pairs.push_back(p1);
pairs.push_back(p2);

const std::string s1;
std::cout << "now we will check \"pairs\": ";
std::cin >> s1;

value(s1);
}
--------------------------------------------------------------------------

basically it is a "vector" of "Pair" to which i added 3 "Pair values".
then i called "value" function which returns a reference to "val"
related to "string" & if "string" is not found in "pairs" then it
simply creates a new "Pair" & adds it to the end of "pairs" .
compilation gave me this error (using "g++ 4.1.1" on BLAG Linux. ):

-------------------------------------------------------------------------
[arnuld@localhost cpp]$ g++ 05_55-references.cpp

05_55-references.cpp: In function 'int main()':
05_55-references.cpp:41: error: no match for 'operator>>' in
'std::cin >> s1'
/usr/lib/gcc/i386-redhat-linux/4.1.1/../../../../include/c++/4.1.1/istream:131:
note: candidates are: std::basic_istream<_CharT, _Traits>&
std::basic_istream<_CharT,

[SNIP]

<_CharT, _Traits>::eek:perator>>(void*&) [with _CharT = char, _Traits =
std::char_traits<char>]
/usr/lib/gcc/i386-redhat-linux/4.1.1/../../../../include/c++/4.1.1/istream:230:
note: std::basic_istream<_CharT, _Traits>&
std::basic_istream<_CharT,
_Traits>::eek:perator>>(std::basic_streambuf<_CharT, _Traits>*) [with
_CharT = char, _Traits = std::char_traits<char>]

[arnuld@localhost cpp]$
 
M

Mike Wahler

arnuld said:
this is the code:

Remarks in-line e
-------------------------------------------------------------------------

#include <iostream>
#include <string>
#include <vector>
[...]

const std::string s1;

Since you used the 'const' qualifer here, that's
your promise that your code will *not* modify
the object 's1'.
std::cout << "now we will check \"pairs\": ";
std::cin >> s1;

.... but here you go trying to modify 's1'. :)
value(s1);
}
[...]
05_55-references.cpp: In function 'int main()':
05_55-references.cpp:41: error: no match for 'operator>>' in
'std::cin >> s1'
/usr/lib/gcc/i386-redhat-linux/4.1.1/../../../../include/c++/4.1.1/istream:131:
note: candidates are: std::basic_istream<_CharT, _Traits>&
std::basic_istream<_CharT,
[...]

what is wrong?

You're trying to modify a const object. Not allowed.
Remove 'const' from your definition of 's1'.

I suspect this was just an 'absent-minded' thing you did, and you
do understand what 'const' means. If not, say so, and we'll explain.

-Mike
 
B

BobR

arnuld wrote in message
this is the code:

-------------------------------------------------------------------------

#include <iostream>
#include <string>
#include <vector>

struct Pair {
std::string name;
double val;
};

std::vector<Pair> pairs;

double& value(const std::string s) {
for(int i=0; i < pairs.size(); i++)
if(s == pairs.name) return pairs.val;

Pair p = {s, -1.1};
pairs.push_back(p); // pair added at the end

return pairs[pairs.size() - 1].val;
}


int main() {

Pair p0 = {"p0", 0.0};
Pair p1 = {"p1", 1.0};
Pair p2 = {"p2", 2.0};

// add to "pairs"
pairs.push_back(p0);
pairs.push_back(p1);
pairs.push_back(p2);

const std::string s1;
std::cout << "now we will check \"pairs\": ";
std::cin >> s1;

value(s1);
}
--------------------------------------------------------------------------

basically it is a "vector" of "Pair" to which i added 3 "Pair values".
then i called "value" function which returns a reference to "val"
related to "string" & if "string" is not found in "pairs" then it
simply creates a new "Pair" & adds it to the end of "pairs" .
compilation gave me this error (using "g++ 4.1.1" on BLAG Linux. ):

-------------------------------------------------------------------------
[arnuld@localhost cpp]$ g++ 05_55-references.cpp

05_55-references.cpp: In function 'int main()':
05_55-references.cpp:41: error: no match for 'operator>>' in
'std::cin >> s1'

[SNIP]
[arnuld@localhost cpp]$
---------------------------------------------------------------------------- ---

what is wrong?


Look at std::string 's1'.
const std::string s1;

The 'const' means you (or anything else in your program) is allowed to change
it. Remove the 'const', and see if it will work.

Usually you would use a const string like:

const std::string name( "This will stay the same for the whole program" );
 
B

BobR

BobR wrote in message ...
Look at std::string 's1'.

The 'const' means you (or anything else in your program) is allowed to change
it. Remove the 'const', and see if it will work.

The 'const' means you (or anything else in your program) is NOT allowed to
change it. Remove the 'const', and see if it will work.
 
A

arnuld

Remarks in-line e

what does that mean?
You're trying to modify a const object. Not allowed.
Remove 'const' from your definition of 's1'.
I suspect this was just an 'absent-minded' thing you did, and you
do understand what 'const' means. If not, say so, and we'll explain.

i do understand what "const" means but that was really absent-minded
thing i did. anyway i removed it & programme compiled & ran but it does
not do what i intended. i also made some minor changes to function
"value" to get output on my terminal:

-----------------------------------------------------------
double& value(const std::string s) {
for(int i=0; i < pairs.size(); i++)
if(s == pairs.name)
{
std::cout << pairs.val << "\n";
return pairs.val;
}

Pair p = {s, -1.1};
pairs.push_back(p); // pair added at the end
std::cout << pairs[pairs.size() - 1].val << "\n";
}
-------------------------------------------------------------


------------------- OUTPUT --------------------------------
[arnuld@localhost cpp]$ ./a.out
now we will check "pairs": p2
2
[arnuld@localhost cpp]$ ./a.out
now we will check "pairs": p9
-1.1
[arnuld@localhost cpp]$
 
A

Amit

Hi arnuld,

If u want to view decimal precision values, u need to set some
properties in cout

try using these statments

cout << showpoint;
cout.precision(2);

Regards.
-Amit Gupta
Mike Wahler wrote:
Remarks in-line e

what does that mean?
You're trying to modify a const object. Not allowed.
Remove 'const' from your definition of 's1'.
I suspect this was just an 'absent-minded' thing you did, and you
do understand what 'const' means. If not, say so, and we'll explain.

i do understand what "const" means but that was really absent-minded
thing i did. anyway i removed it & programme compiled & ran but it does
not do what i intended. i also made some minor changes to function
"value" to get output on my terminal:

-----------------------------------------------------------
double& value(const std::string s) {
for(int i=0; i < pairs.size(); i++)
if(s == pairs.name)
{
std::cout << pairs.val << "\n";
return pairs.val;
}

Pair p = {s, -1.1};
pairs.push_back(p); // pair added at the end
std::cout << pairs[pairs.size() - 1].val << "\n";
}
-------------------------------------------------------------


------------------- OUTPUT --------------------------------
[arnuld@localhost cpp]$ ./a.out
now we will check "pairs": p2
2
[arnuld@localhost cpp]$ ./a.out
now we will check "pairs": p9
-1.1
[arnuld@localhost cpp]$
 
A

arnuld

Amit said:
Hi arnuld,

If u want to view decimal precision values, u need to set some
properties in cout

try using these statments

cout << showpoint;
cout.precision(2);

1st, do not top-post.

2nd, you did not get my point. my input was "2.0" ( Pair p2 = { "p2",
2.0} )

i wanted to know why compiler changed it to /int 2/. on the contrary,
if i input / Pair p3 = { "p3", 3.3} / i get 3.3 as output. so this time
compiler did not change it to /int 3/. why so?
 
B

BobR

Amit wrote in message
Hi arnuld,

If u want to view decimal precision values, u need to set some
properties in cout

try using these statments

cout << showpoint;
cout.precision(2);

Regards.
-Amit Gupta

Nice top-post. Been takeing lessons long?


(http://www.parashift.com/c++-faq-lite/how-to-post.html#faq-5.4).

A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
 
B

BobR

arnuld wrote in message ...
1st, do not top-post.

2nd, you did not get my point. my input was "2.0" ( Pair p2 = { "p2",
2.0} )

i wanted to know why compiler changed it to /int 2/. on the contrary,
if i input / Pair p3 = { "p3", 3.3} / i get 3.3 as output. so this time
compiler did not change it to /int 3/. why so?

What evidence do you have that it is an 'int'? What you see in the output may
be very different than what the variable holds.

Put this in main() and try it:

{
using std::cout;
cout.setf( std::ios_base::fixed );
cout.precision( 20 );
double d1 = 8.126e9;
cout <<"d1 = "<< d1 <<std::endl;

cout.setf( std::ios_base::scientific );
cout.precision(2);
cout <<"d1 = "<< d1 <<std::endl;
cout.precision(6);
cout <<"d1 = "<< d1 <<std::endl;
cout.setf( std::ios_base::fixed );
}
 
A

arnuld

BobR said:
What evidence do you have that it is an 'int'?

becuase i get plain 2 ( not 2.0 )
What you see in the output may
be very different than what the variable holds.

then how will i get the desired output? or how will i know what the
variable holds?
2nd, is it really necessary, practically, to know answers to these 2
questions i have asked?

BTW, what is the "type" of "2" (the output i got)?
Put this in main() and try it:

{
using std::cout;
cout.setf( std::ios_base::fixed );
cout.precision( 20 );
double d1 = 8.126e9;
cout <<"d1 = "<< d1 <<std::endl;

cout.setf( std::ios_base::scientific );
cout.precision(2);
cout <<"d1 = "<< d1 <<std::endl;
cout.precision(6);
cout <<"d1 = "<< d1 <<std::endl;
cout.setf( std::ios_base::fixed );
}

this is the output from "g++ 4.1.1 on BLAG Linux":

-------------------------------------------------------------------
[arnuld@localhost ~]$ ./a.out
d1 = 8126000000.00000000000000000000
d1 = 8.1e+09
d1 = 8.126e+09
[arnuld@localhost ~]$
 
A

Alf P. Steinbach

* arnuld:
becuase i get plain 2 ( not 2.0 )

The presentation does not tell you anything about the type.

Also, in the case of floating point it only tells you the value
/approximately/, because the internal representation of a floating point
number is finite, and thus cannot represent all numbers exactly.
 
A

arnuld

Alf said:
The presentation does not tell you anything about the type.
:-(

Also, in the case of floating point it only tells you the value
/approximately/, because the internal representation of a floating point
number is finite, and thus cannot represent all numbers exactly.

OK, now i got it

thanks Alf
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

these signatures taught me how irritating a top-post is. i never did
that since i read your signatures some months ago. thanks for putting
them.

[OT]
BTW, how can i put my signatures in my post so that they appear
everytime automatically. i googled for it, i have signatures in my
gmail but i am not able to know how to do the same at newsgroups.
[/OT]
 
B

BobR

arnuld wrote in message ...
BobR wrote:

BTW, what is the "type" of "2" (the output i got)?
char.


this is the output from "g++ 4.1.1 on BLAG Linux":
-------------------------------------------------------------------
[arnuld@localhost ~]$ ./a.out
d1 = 8126000000.00000000000000000000
d1 = 8.1e+09
d1 = 8.126e+09
[arnuld@localhost ~]$

Doc: "The GNU C++ Iostream Library"
"Choices in formatting"
"Changing stream properties using manipulators"

'setf()' is Set Flags.
 
D

Default User

arnuld wrote:

[OT]
BTW, how can i put my signatures in my post so that they appear
everytime automatically. i googled for it, i have signatures in my
gmail but i am not able to know how to do the same at newsgroups.
[/OT]


Google Groups does not have that capability, along with many other
features of a real newsreading system. A possible remedy is to get
access to a genuine newsfeed, and then use the newsreader (there are
many available) that has the features you desire.

Your ISP may provide access to a news server. If not, an inexpensive
alternative is http://news.individual.net (the German server), which
costs 10 euro per year (about $13 US currently) for access to text
newsgroups.

There are some free ones. Generally their reliability and retention
aren't as good as NIN, but they are worth looking into if you don't
want to pay.




Brian
 

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