sizeof question

Code:

char *f;
f = "123456789";
cout << "size of f is " << sizeof(f) << endl;

Why is this 4? Everytime I compile it, its 4.
No matter what I put the value of f to be, it says 4.

I would really like to know how I get the size of a char array like this, any reply appreciated.

Dag

Why is this 4?

Probably because sizeof(int) on your system is 4.

I would really like to know how I get the size of a char array like this

Use strlen() from <cstring> ?

Use strlen() from <cstring> ?

Thanks you!, that is just what I wanted

Test this

Code:

                char first[]="123456789";
	char* second = "123456789";
	cout<<sizeof(first)<<endl;
	cout<<sizeof(second);

And see what you can come up with!

Because you're having 32-bit machine that means that every address is 4 byte wide ant that means sizeof(void*) is equal like sizeof(char*) i.e. 4 bytes.
And see what is the difference between first and second.

Think about it a little.

Originally Posted by laserlight

Probably because sizeof(int) on your system is 4.

Actually, it's because the sizeof the pointer is 4 bytes.

Code:

char *f;
f = "123456789";
cout << "size of f is " << sizeof(f) << endl;

Is that legal? First of all, f is uninitalized. You'll also need some memory to hold your string. Second, I think you really want strcpy.

Code:

char *f;
f = new char[20];
strcpy(f,"123456789");
cout << "size of f is " << sizeof(f) << endl;
cout << f << endl;
delete[] f;

Originally Posted by pianorain

Code:

char *f;
f = "123456789";
cout << "size of f is " << sizeof(f) << endl;

Is that legal? First of all, f is uninitalized.

Sure it's initialized, it's initialized to the address of the string-literal. It should however probably be a const char*.

Actually, it's because the sizeof the pointer is 4 bytes.

hmm... I had the impression that the sizeof of any pointer would be the same as sizeof of int, since a pointer is an integer.

>>since a pointer is an integer.
Not sure about this one, it can be interpreted as an integral type but I seem to remember hearing somewhere that it is NOT an integer.

>>but I seem to remember hearing somewhere that it is NOT an integer.

Like on a 64-bit machine where pointers are... 64 bits. ints and pointers are not gaurenteed to be the same size.

char *f;
f = "123456789";
cout << "size of f is " << sizeof(f) << endl;

Why is this 4? Everytime I compile it, its 4.

...because pointers like f store addresses. "1234567890" is a 'string literal' and it is placed in memory somewhere, and the following assignment:

f = "123456789";

assigns the address of the string literal to f. An address on your system is 4 bytes long, and no matter how long the string is, it's address in memory--which is the address of the first character of the string--will always be 4 bytes long.

wow, lots of replies!
I would like to thank all who explained this sizeof thing also! I think I understand it a bit more now...!
greatly appreciated!

Dag