C++ char* problems [closed] - c++

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#include<iostream>
using namespace std;
int main()
{
char *name="Siva",*str;
for(int i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
cout<<str;
return 0;
}
The first program gives output Sivaœ> i.e siva with some garbage values....
But the second program show segmentation fault...Please help me to find out
exact answer...
#include<iostream>
using namespace std;
int main()
{
int i=0;
char *name="Siva",*str;
for(i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
cout<<str;
return 0;
}

Answer 1

char *name="Siva",*str;
for(int i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
str is a pointer, but it doesn't yet point to anything.
Since you're in C++, you should be using std::string:
#include<iostream>
#include <string>
using namespace std;
int main()
{
char *name="Siva";
std::string str;
for(int i=0;i<strlen(name);i++)
{
str += name[i];
}
cout<<str;
return 0;
}
Even better, get rid of the hand-written loop:
#include <algorithm>
int main()
{
char *name="Siva";
std::string str;
std::copy (name, name + strlen (name), std::back_inserter (str));
cout<<str;
return 0;
}
Better even still, there's no reason in this particular example why you need to do any of that at all:
char* name = "Silva";
std::string str = name;
cout << str;
By the way, string literals in C++ are inherently const:
const char* name = "Silva";
If you really must use a char*, first I would strongly question why, and then I would tell you to do this:
int main()
{
const char *name="Siva";
char* str = new char [strlen (name)+1]; // +1 for the null terminator
strcpy (name, str);
cout << str;
delete [] str;
return 0;
}
I would even more strongly question your need to copy it byte-by-byte:
int main()
{
const char *name="Siva";
char* str = new char [strlen (name)+1]; // +1 for the null terminator
for (size_t i = 0; i < strlen (name); ++i )
str [i] = name [i];
str [strlen (name)] = '\0';
cout << str;
delete [] str;
return 0;
}

Answer 2

You have undefined behaviour here:
str[i]=name[i];
str has not been initialized to anything. You are writing to places you shouldn't.

Answer 3

There are two problems with this.
The pointer str doesn't point to allocated memory, so writing through it is undefined behavior.
Even if it did point to valid memory, you're not writing the correct amount of data. When copying a string, you need to copy the 0 byte at the end which marks the end of the string; so the upper limit of your loop should bt strlen(name) + 1. Or you could use a library method like strdup() instead of your own for loop.
The reason the "working" version prints some garbage characters is that there's no 0 at the end of the copied string to tell iostreams to stop printing. The reason the "working" one doesn't crash, and the other one does, is pure dumb luck: the garbage in str, by chance, points to memory you're allowed to write to, while in the crashing program, it points to memory you're not allowed to write to. Simple as that.

Answer 4

It is because you have no memory allocated for str. (it will cause an undefined behavior)
You can mix that by using a merory allocation function like in this example :
#include<iostream>
using namespace std;
int main()
{
char *name="Siva",*str;
// Allocate memory with malloc
str = (char*)malloc( (strlen(name)+1) * sizeof(char) );
for(int i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
str[strlen(name)] = 0;
cout<<str;
// Free the allocated memory
free(str);
return 0;
}
As you are using c++, you can do :
#include<iostream>
using namespace std;
int main()
{
char *name="Siva",*str;
// Allocate memory with new
str = new char[strlen(name) + 1];
for(int i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
str[strlen(name)] = 0;
cout<<str;
// Free the allocated memory
delete []str;
return 0;
}
EDIT :
The reason you have some weird caracters at the end of your ouput is that because you string is not terminated with a '\0', it will continue to print it. (This will occur only if you don't have a segmentation fault )

Answer 5

There are a couple problems with your code.
Firstly, *str is not allocated, so it starts off by pointing to whatever bit of memory the pointer value happens to start off as.
Secondly, strlen() returns the length of the string excluding the terminating null character. So what you are doing is copying all the values of name into some random bit of memory, not terminating it, then telling the system to print that off, which could be any length.

Answer 6

str[i]=name[i]; is illegal, causes Undefined behavior, because you have not allocated memory for str.
Before for loop allocate memory for destination string str:
str = malloc (strlen(name) + 1 );
Also you forgetting string termination, after for-loop add terminate string str[i] = '\0';
Undefined behavior refers to computer code whose behavior is unpredictable.
You code should be:
char *name="Siva", *str;
str = malloc (strlen(name) + 1 ); // mistake
for(int i=0;i<strlen(name);i++)
{
str[i]=name[i];
}
str[i] = '\0'; // forgetting
To understand further you can read this answer: strcat() implementation works but causes a core dump at the end

Answer 7

Your problem is using char arrays and pointers to represent strings in language which has proper string type.
#include <iostream>
#include <string>
using namespace std;
int main()
{
string name = "Siva", str;
str = name;
cout << str;
return 0;
}

Answer 8

Problem is with str[i]=name[i] you must know that C++ does not care for memory leaks
as Java or some others. So you must allocate memory to a variable or pointer in order to avoid these problems. Already there are number of answers, you can also try
str=new char[strlen(name)+1];
and do not forget to terminate the char array with null when you done copy. In this case
str[strlen(name)]='\0';