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Asked: 2026-05-26T19:20:46+00:00

Here is what my book says: char *p=Hello; //pointer is variable, string is constant

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Here is what my book says:

char *p="Hello"; //pointer is variable, string is constant
*p='M'; // error
p="bye"; // works

Well, in C my second line does not give me any error, Neither in C++.

I am using Turbo C++ on windows 7. Now is the above thing true if i try in gcc or something else.

Also on the similar lines if the above code is correctly interpreted by the book, 

#include<iostream.h>
void display(char*);
void display(const char*);
void main()
{
char* ch1="hello";
const char *ch2="bye";
display(ch1);
display(ch2);
}

void display(char* p)
{
cout << p << endl;
}

void display(const char* p)
{
cout << p << endl;
}

Now is my book considering char* and const char* the same because if it is then the above code shall not work, since arguments will be the same?

(Though i get the output Hello bye on turbo +windows.)

Which one is correct?

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1 Answer

  1. Editorial Team

    The language specification is not a promise that compilers make to you, but a mutual contract that you and compilers both sign onto. (Metaphorically speaking, of course. Obviously it’s not a literally legally binding contract.) If you violate the contract by writing *p='M';, then you’ve triggered “undefined behavior”, and you can’t expect any specific behavior from the compiler: maybe it will be strict about it and give you a compile error, maybe it will just go wonky at run-time . . . you didn’t hold up your end of the bargain, and it’s allowed to do literally whatever it wants now. See also: http://catb.org/jargon/html/N/nasal-demons.html.

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