Inserting value and printing the value

[BABU Prasad G]

Dear friends ,
Please could you help me to get out of this problem.

I want to assign a value and print it again. but it is giving error :
as
problem1.cpp:33: error: expected initializer before ‘srl1’

program :
#include<iostream>
#include<vector>
struct linkparam {

int value;
};
typedef std::vector <linkparam> link1;

struct link_status_rsp {
link1 ll;
};
struct status_rsp {

link_status_rsp rsp;
};

typedef std::vector<status_rsp> srl;


int main()
{
link_status_rsp lsr;

linkparam lp;
lp.value =90;

//link1.push_back(lp);
link_status_rsp lsr;
//lsr.l1 =
status_rsp sr;
sr.rsp.ll.push_back(lp);
//typedef std::vector<status_rsp> srl;
srl srl1
srl1.insert(srl.end(), sr);

//srl.push_back(sr);
}

and also please tell me how to print the value of vector "srl1"


Best Regards
BABU

[codekiddy]

program :
#include<iostream>
#include<vector>
struct linkparam {

int value;
};
typedef std::vector <linkparam> link1;

struct link_status_rsp {
link1 ll;
};
struct status_rsp {

link_status_rsp rsp;
};

typedef std::vector<status_rsp> srl;


int main()
{
link_status_rsp lsr;

linkparam lp;
lp.value =90;

//link1.push_back(lp);
link_status_rsp lsr;
//lsr.l1 = 
status_rsp sr;
sr.rsp.ll.push_back(lp);
//typedef std::vector<status_rsp> srl;
srl srl1
srl1.insert(srl.end(), sr);

//srl.push_back(sr);
}

I've just copyed.
what does this mean: ?
sr.rsp.ll.push_back(lp);

you're missing semicolon after srl1

[BABU Prasad G]

Dear,
sr.rsp.ll.push_back(lp); ......According to me.

the variable "value" in linkparam structure should be passed to status_rsp structure for that, I used like that .

and even after corrected the semicolon after srl1. now the error is :


problem1.cpp:33: error: expected primary-expression before ‘.’ token

[BABU Prasad G]

please help me get out of this problem and also tell me how to print the content of srl1 vector..

[codekiddy]

link_status_rsp lsr; allready defined on line 23

[BABU Prasad G]

yes after corrected .Even though it is giving same error as earlier.. please

[codekiddy]

also tell me how to print the content of srl1 vector

1. create 2x const_iterator
2. create for loop
3. print values trought iterator to the screen within for loop.

for example:

typedef std::vector<status_rsp> srl;
srl::const_iterator iter = srl.begin();
srl::const_iterator iter2 = srl.end();
for(;iter != iter2; ++iter)
     cout << iter->ll.value;

[BABU Prasad G]

Dear friend,

It is giving error while assign the value time itself. I ran in code in "codepad.org" also but same error

[BABU Prasad G]

Thanks for fast and helpful suggestions

[BABU Prasad G]

could you tell me please how to assign value to " value " variable and through srl1 vector..

[codekiddy]

srl1::iterator iter = srl1.begin();  //not const this time !
iter->ll.value = 30;
++iter;
iter->ll.value = 99;
++++iter;  // 2 time ahead;
iter->ll.value = 5;

for(iter; iter != srl.end(); ++iter)
     iter->ll.value = "value";

[BABU Prasad G]

Dear friend , still error is the

#include<iostream>

#include<vector>
struct linkparam {

int value;
};
typedef std::vector <linkparam> link1;

struct link_status_rsp {
link1 ll;
};
struct status_rsp {

link_status_rsp rsp;
};

typedef std::vector<status_rsp> srl;


int main()
{


linkparam lp;
//lp.value =90;

link_status_rsp lsr;

status_rsp sr;
srl srl1;
std::vector<status_rsp>::iterator iter = srl1.begin(); //not const this time !
lp.value = 30;
iter->rsp.ll.push_back(lp);
++iter;
lp.value = 60;
iter->rsp.ll.push_back(lp);
++++iter; // 2 time ahead;
lp.value = 90;
iter->rsp.ll.push_back(lp);

for(iter; iter != srl->end(); ++iter)
std::cout<< iter->rsp.ll.value = "value"<<endl;





}
please could you help me to get out of this problem..

program: after little modification


error:: n function 'int main()':
Line 41: error: expected primary-expression before '->' token
compilation terminated due to -Wfatal-errors.

[BABU Prasad G]

This is the sample and similar and part of the code of my program . So in my program when I ran the code It is giving the error as


Un able to match the operator << in { some big message is coming }

please tell me either please atleast .

[BABU Prasad G]

This error is showing at printing line ie std:: cout << ........

[codekiddy]

Ok BABU, here is your code!

#include<vector>
#include<iostream>
using namespace std;

struct linkparam {
	int value;
};

typedef vector<linkparam> link1;
struct link_status_rsp {
	link1 ll;
};

struct status_rsp {
	link_status_rsp rsp;
};
typedef vector<status_rsp> srl;

int main()
{
	//Destiantion
	linkparam lp;
	lp.value =90;

	//Node1
	status_rsp sr;
	sr.rsp.ll.push_back(lp);

	
	//Node2
	srl srl1;
	srl1.push_back(sr);

	//Setup Access 
	link1::iterator iter2 = srl1.begin()->rsp.ll.begin();

	//list Destination value
	cout << "A vlue is: " << iter2->value;

	cin.ignore();
	return 0;
}

You old me a bottle of wiskey for that :D

[bluecoder]

int main()
{
        linkparam lp;
	lp.value =90;
	link_status_rsp lsr;
	status_rsp sr;
	sr.rsp.ll.push_back( lp );

	srl srl1;
	srl1.push_back( sr);
        return 0;
}

[codekiddy]

@bluecoder

Your code is exactly the same as mine, except it does not print a value and does not contain comment's.

so what's your point exactly? :D

are you trying to say that your code is more efficient than mine?

[bluecoder]

Nope codekiddy .. actually i did not see your code and pasted mine ..sorry for that i should have looked at it . ..thanks for the code .. and sorry for not looking at your code first .. have a nice day :)

[BABU Prasad G]

Dear friends,
Please look my code and when I trying to print the values of structure "status_rsp" using insertion operator overloading but it is giving "error as " no match operator << .

how to print value by using insertion operator overloading , because here ie in this sample code without overloading also printing but this is the piece of or sample of code , so once I executed there it is giving error as " no match operator for <<..

please any one tell me how to get out of this problem.


code:


#include<vector>
#include<iostream>
using namespace std;

struct linkparam {
int value;
};

typedef vector<linkparam> link1;
struct link_status_rsp {
link1 ll;
};

struct status_rsp {
link_status_rsp rsp;
friend std::ostream& operator<<(std::ostream& os , const status_rsp& sr)
{
os << "the value is " << sr.rsp;
}
};
typedef vector<status_rsp> srl;

int main()
{
//Destiantion
linkparam lp;
lp.value =90;

//Node1
status_rsp sr;
sr.rsp.ll.push_back(lp);


//Node2
srl srl1;
srl1.push_back(sr);

//Setup Access
link1::iterator iter2 = srl1.begin()->rsp.ll.begin();

//list Destination value
cout << "A vlue is: " << iter2->value;

cin.ignore();
return 0;
}

[BABU Prasad G]

please urgent , any can solve this pls..