Very basic question

[chizarlicious]

This is pretty embarrassing... I've been coding in C++ off and on for 5 years now. I'm trying to run the following routine but getting a seg failt when I try to assign the 'f' . It doesn't matter if I try to assign it to str[0] or str[3] or if I try to assign 'o' or '7'. I also get the same problem if I declare str as char str[5].
Help?

int main(){
char* str;
str = new char[5];
str = "test";
cout << str << endl;
str[1] = 'f'; // causes seg fault
cout << str << endl;
return 0;
}

[GCDEF]

"test" is a string literal.
str = "test"; assigns the address of the string literal to str.
str[1] = 'f'; tries to modify the string literal which you can't do.

you need

strcpy(str, "test");

to assign a value to str.

[Lindley]

If you are really interested in programming C++ rather than C, then consider declaring str as a std::string rather than a char array/pointer.

[Muthuveerappan]

Yes GCDEF is correct.

Just thought i would add some points (correct me if i am wrong):
1) C-Strings

C-Strings are treated as an array of characters ending with a null character '\0'
C-String pointer would contain the starting address of the C-String (length of the string is determined based on the occurrence of the null character '\0')

2) I suppose a warning should have been thrown for the following statement (warning messages tend to vary based on the compiler):

str = "test";

GCC warning message:

warning: deprecated conversion from string constant to ‘char*’

It means that str is declared as char* but is being assigned the address of constant char

Therefore a char* variable is being assigned a const char* value which is not correct.

"test" is a literal whose value can't be changed and therefore the address of this literal would be of the const char*

This warning would not be thrown, if str was declared as a const char*
and if str was declared as const* then str[1] = 'f' wouldn't be allowed

3) As GCDEF has stated, strcpy is what you intended to use

strcpy will copy the the characters that "test" to the memory location that str points to.

4) Memory for the literal is allocated and is available till the end of the program

So printing the address of the literal "test" several times in the same program would be same address till the end of the program

5) print the address of str to understand better before and after the statement str = test

Note - This program still will throw a warning (at least in GCC)

#include <iostream>
using std :: cout;
using std :: endl;

int main()
{
    char* str;
    str = new char[5];
    cout << "str = " << static_cast<void*>(str) << endl;
        
    str = "test";
    cout << "str = " << static_cast<void*>(str) << endl;
        
    cout << str << endl;
    str[1] = 'f'; // causes seg fault
    cout << str << endl;
    return 0;
}

Note - static_cast is used so that cout doesn't print the value of the C-string str point to and instead prints the address of the C-String