How to access the std::string type data?

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CString s=strLine;
int size=s.GetLength();
char* NotesStr;
std::vector<char> notesChar;
int c=0;
for(int n=0;n<size;n++)
{

         while((s[n]!='.')&&s[n]!='T')
	{
		notesChar.push_back(s[n]);
		c++;
		n++;
	}
	if(s[n]=='.')
	{
		std::string NotesStr(notesChar.begin(),notesChar.end());
		for(int p=0;p<c;p++)
		{
			notesChar.pop_back();
						
		}
		c=0;
	}
	else if(s[n]=='T')
	{
		std::string NotesStr(notesChar.begin(),notesChar.end());
		if(NotesStr=='1O')//error comes here
		{
			MessageBox("Hi");
		}
		for(int p=0;p<=c;p++)
		{
			notesChar.pop_back();
						
		}
		c=0;
		break;
	}
}

The above code is giving my the following error
error C2678: binary "==": no operator found which takes a left-hand operand of type std::string(or there is no acceptable conversion)

Posted 6-Mar-12 23:04pm

chaiein

Updated 6-Mar-12 23:07pm

v3

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4 solutions

Solution 2

what if you change it by:

C++

if(NotesStr=="1O")

Posted 6-Mar-12 23:39pm

Schehaider_Aymen

Comments

chaiein 7-Mar-12 6:09am

no even that cause error but different error

chaiein 7-Mar-12 6:09am

std::string t="1F";
if(NotesStr.compare(1,t.length(),t))
{
MessageBox("Hi");
}

chaiein 7-Mar-12 6:09am

The above code works

Solution 1

The reason for the compiler error is that there is no operator for comparing two values of type std::string and char. If your purpose really is to compare two full strings, then you should convert the operand on the right side to std::string as well, as there is in fact a defined operator == for comparing two operands of type std::string.

That said, your right hand side operand uses single hyphens, indicating a single character, not a string. And it isn't even a valid character.

Did you mean a single character with the decimal (or hexadecimal?) value 10? If so, you cannot compare an entire string with a single character, so what was your intention?

Or did you mean the string "10"? If so, the comparison will always be false, as the string you compare to does not contain a 'T', and that is the condition you verified just before. So what did you mean to test?

Posted 6-Mar-12 23:36pm

Stefan_Lang

Comments

Schehaider_Aymen 7-Mar-12 5:47am

I don't think that the no use of 'T' will cause an issue, the code below works fine :

<pre lang="c++">
string s2;
s2 ="Test string";

if (s2 == "Test string")
cout << "YES" << endl;

</pre>

Stefan_Lang 8-Mar-12 3:56am

That thing about 'T' was just a misunderstanding on my part, I thought the preceding if condition was referring to the same string, which it didn't. So that last paragraph of mine is meaningless.

chaiein 7-Mar-12 6:04am

1O is not ten its one and 'O' Alphabet

Schehaider_Aymen 7-Mar-12 6:09am

the prob is not 1 and which litteral, if you wanna comapre the string to another and this latter contains just one alphabet so use the ' but if it contains 2 or more alphabets then you ll need the ".

chaiein 8-Mar-12 3:35am

Thanks for the reply:)
I had to access the data to compare. I got the solution as below
std::string t="1O";
if(NotesStr.compare(1,t.length(),t))

Stefan_Lang 8-Mar-12 3:55am

It seems I misunderstood part of your posting: ignore the part in my answer referring to 'T' - that was based on my misunderstanding that the preceding if condition tested the same string, which it didn't.

Anyway, good to know you could solve your problem.

Solution 4

you can see two variables with same name "NotesStr" in that code block, Try changing the name of std::string NotesStr(notesChar.begin(),notesChar.end());
to some other name.

Also in order to compare string you need to use double quotes instead of single quotes.
string test = "whoami";
if (test == "whoami") printf(...) like that.

jkchan
http://cgmath.blogspot.com

Posted 7-Mar-12 0:08am

jk chan

Comments

chaiein 8-Mar-12 3:35am

Thanks for the reply:)
I had to access the data to compare. I got the solution as below
std::string t="1O";
if(NotesStr.compare(1,t.length(),t))

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chaiein · Accepted Answer · 2012-03-07T00:08:00

Solution 3

C#

std::string t="1F";
 if(NotesStr.compare(1,t.length(),t))
 {
       MessageBox("Hi");
 }

This works:)

Posted 7-Mar-12 0:08am

chaiein