Finding letters in a vector of strings

[pel1993]

hi i need to find repeating letters in a string of vectors. what am i do wrong here i am getting the memory address (some random number) but not like the number of times 'cg' occur.

int count=0;
    for (string::iterator it = dna[0].begin(), endOfString = dna[0].end(); it != endOfString; it++)
    if ( *it ==  'cg')
      count++;

[tfityo]

You can you find method:

dna[0].find("cg");

Read full description of it: http://www.cplusplus.com/reference/string/string/find/

[andywestken]

1. The string iterator will process the string on a per-element basis (here, elements are chars). So *it is a single char ('c' or 'g' or ...). So your if-test is attempting to test a single char against a pair of them.

2. Strings in C and C++ are in "s. So "cg" makes sense, as a string. But not 'cg'

What 'cg' is, in fact, a const integer.

char c = 'cg';

1>c:\source\test.cpp(81) : warning C4305: 'initializing' : truncation from 'int' to 'char'
1>c:\source\test.cpp(81) : warning C4309: 'initializing' : truncation of constant value

char* p = 'pqrs';

1>c:\source\test.cpp(101) : error C2440: 'initializing' : cannot convert from 'int' to 'char *'

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

int main()
{
	int  i = 'abcd'; // allowed up to 4 chars (for 32-bit int)

	cout << 'pqrs' << endl;

	cout << hex;
	cout << i << endl;
	cout << int('a') << int('b') << int('c') << int('d') << endl;

	return 0;
}

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61626364

tfityo's suggestion -- the find method -- is going to be the easiest approach (making use of the second, optional "pos" parameter, which says where to start the search from)

Andy