Detecting a word

[G26pX9L8]

Hello,

ive a little question. how do I know if a given word in a sentence, so I then can perform an action?
Example:
this is an example
check the word:
example
now if example is found i what to run some code. Does anybody know how to solve this?

[Serpent]

This is as far as I got, sorry I know it doesn't work. At least the concept is right! :D If you or anyone would make it work that would be great!

#include <windows.h>
#include <winuser.h>
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <string>

using namespace std;

string input;
char convert[50];
char word1[50];
char word2[50];
char word3[50];
char space[]=" ";
bool go = false;

int main()
{
	input = "he man example";
	strcpy_s(convert, input.c_str());
		for (int i=0; i<50; i++)
		{
			while (convert[i] != space[0])
			{
				go=true;
				word1[i]=convert[i];
				while (go)
				{
					i++;
					go=false;
				}
			}
			while (convert[i] != space[0])
			{
				go=true;
				word2[i]=convert[i];
				while (go)
				{
					i++;
					go=false;
				}
			}
			while (convert[i] != space[0])
			{
				go=true;
				word3[i]=convert[i];
				while (go)
				{
					i++;
					go=false;
				}
			}
		}
	if (word1 == "example")
		cout << "word1!\n";
	if (word2 == "example")
		cout << "word2!\n";
	if (word3 == "example")
		cout << "word3!\n";
	return 0;
}

[filipe]

Here is how you could do it:

	std::string str;
	getline(std::cin, str);

	for(std::string::size_type i = 0; i < str.size(); ++i)
	{
		if(!isalpha(str[i]))
		{
			str[i] = ' ';
		}
	}

First, read a whole line into a std::string and get rid of any character that isn't a letter. This will keep you from reading something like "example." as if it were a word. Do notice that this prevents the program from reading words with the character '-' in them, so adjust as needed.

	std::stringstream buffer;
	buffer << str;
	while(buffer)
	{
		std::string s;
		buffer >> s;
		if(s == "example")
		{
			std::cout << s << std::endl;
			break;
		}
	}

Then, parse the buffer using a std::stringstream and check for the word you're looking for. In this case, we simply output the word and break from the loop, but in a real program this should really be inside a function, something like bool f(std::string line, std::string word) that would return if it found the word.

One last thing: this really has nothing to do with Windows so I guess it doesn't fit the Windows programming forum.

[D80DSL3A]

This solution is based on the use of a standard string function strstr( char* s1, char* s2) which returns a pointer to the location of s2 within s1. It is a console app (not a windows app).
I have initialized a char* to NULL here. If it stays NULL it is because the word was not found within the sentence. Note: As written the sentence length is limited to 49 characters so adjust if needed for a longer one.

#include <iostream>// for use of cin, cout
#include <cstdio>// for use of gets(), scanf()
#include <cstring>// for use of string functions strcpy(), strcmp(), strlen(), strcat(), strstr(), etc.

using namespace std;


int main()
{
	char bigString[50];
	char word[50];	
	char* pLocation = NULL;
	char a = 'n';

	do  // allows program to repeat without restart
	{		
		cout << "enter big string:";
		gets_s(bigString);
		cout << "enter a word:";
		gets_s(word);
	
		pLocation = strstr(bigString, word);// this string function returns pointer to 2nd string within 1st
		if( pLocation == NULL )
			cout << word << " is not in the sentence: " << bigString << endl;
		else
			cout << word << " is in the sentence: " << bigString << endl;
	
		cout << "repeat (y/n)?" << endl;
		scanf_s("%c", &a);
		_flushall();
               pLocation = NULL;// reset for next loop
	}while( a=='y');// if anything other than 'y' is entered program will terminate

	return 0;
}

Are you trying to create the functionality of the strstr() function yourself? If so then this "solution" misses the point of your exercise.

[G26pX9L8]

OK, tnx all. I'm going to work tonight and then I will let you know more. And it is true that this is not the proper forum.