How to use memset while handling strings in C++?

Vlad has helpfully answered the first part of your question, but I feel like the second part could be explained a little more intuitively:

As others have mentioned, 't' is a character whereas "t" is a string, and strings have a null terminator at the end. This makes "t" an array of not one but two characters - ['t', '\0']! This makes memset's error more intuitive - it can coerce a single char to an int easily enough, but it chokes when it's given an array of chars. Just like in Python, int(['t', '\0']) (or ord(['t', '\0'])) doesn't compute.

Error when using single quotes 't' This prints extra characters.

That's because you overwrote the null terminator.

The terminator is part of the array's size (an array is not magic), though it's not part of the logical string size.

So, I think you meant:

memset(str, 't', strlen(str));
//               ^^^^^^

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Error when using "t" with double quotes

Completely different thing. You told the computer to set every character in the string, to a string. Doesn't make sense; won't compile.

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How to use memset in C++?

Don't.

Either use the type-safe std::fill, in combination with std::begin and std::end:

std::fill(std::begin(str), std::end(str)-1, 't');

(If you're worried about performance, don't be: this will just delegate to memset where possible via template specialisation, optimisation not required, without sacrificing type-safety; example here in libstdc++.)

Or just a std::string to begin with. 😊

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I was learning the fuction memset in C++ from https://www.geeksforgeeks.org/memset-in-cpp/ where the example is given as below

Don't attempt to learn C++ from random websites. Get yourself a good book instead.

This declaration

char str[] = "geeksforgeeks";

declares a character array that contains a string that is a sequence of characters including the terminating zero symbol '\0'.

You can imagine the declaration the following equivalent way

char str[] = 
{ 
    'g', 'e', 'e', 'k', 's', 'f', 'o', 'r', 'g', 'e', 'e', 'k', 's', '\0'
};

This call of the function memset

memset(str, 't', sizeof(str));

overrides all characters of the array including the terminating zero.

So the next statement

cout << str << endl;

results in undefined behavior because it outputs characters until the terminating zero is encountered.

You could write instead

#include <iostream>
#include <cstring>

int main()
{
    char str[] = "geeksforgeeks";

    std::memset( str, 't', sizeof( str ) - 1 );
    
    std::cout << str << '\n';
}

Or the following way

#include <iostream>
#include <cstring>

int main()
{
    char str[] = "geeksforgeeks";

    std::memset( str, 't', std::strlen( str ) );
    
    std::cout << str << '\n';
}

That is keeping the terminating zero unchanged in the array.

If you want to override all characters of the array including the terminating zero, then you should substitute this statement

std::cout << str << '\n';

for this statement

std::cout.write( str, sizeof( str ) ) << '\n';

as it is shown in the program below because the array now does not contain a string.

#include <iostream>
#include <cstring>

int main()
{
    char str[] = "geeksforgeeks";

    std::memset( str, 't', sizeof( str ) );
    
    std::cout.write( str, sizeof( str ) ) << '\n';
}

As for this call

memset(str, "t", sizeof(str));

then the type of the second argument (that is the type const char *) does not correspond to the type of the second function parameter that has the type int. See the declaration of the function

void * memset ( void * ptr, int value, size_t num );

Thus the compiler issues an error message.

Apart from character arrays (that are used very often even in C++) you can use also the standard class std::string (or std::basic_string) that simulates strings.

In this case there is no need to use the standard C function memset to fill a string with a single character. The simplest way to do this is the following

#include <iostream>
#include <string>

int main()
{
    std::string s( "geeksforgeeks" );
    
    s.assign( s.length(), 't' );
    
    std::cout << s << '\n';
}

Another way is to use the standard algorithm std::fill or std::fill_n declared in the header <algorithm>. For example

#include <iostream>
#include <string>
#include <iterator>
#include <algorithm>

int main()
{
    std::string s( "geeksforgeeks" );
    
    std::fill( std::begin( s ), std::end( s ), 't' );
    
    std::cout << s << '\n';
}

or

#include <iostream>
#include <string>
#include <iterator>
#include <algorithm>

int main()
{
    std::string s( "geeksforgeeks" );
    
    std::fill_n( std::begin( s ), s.length(), 't' );
    
    std::cout << s << '\n';
}

You even can use the method replace of the class std::string one of the following ways

#include <iostream>
#include <string>

int main()
{
    std::string s( "geeksforgeeks" );
    
    s.replace( 0, s.length(), s.length(), 't' );
    
    std::cout << s << '\n';
}

Or

#include <iostream>
#include <string>

int main()
{
    std::string s( "geeksforgeeks" );
    
    s.replace( std::begin( s ), std::end( s ), s.length(), 't' );
    
    std::cout << s << '\n';
}