Why is this 4? Everytime I compile it, its 4.Code:char *f; f = "123456789"; cout << "size of f is " << sizeof(f) << endl;
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
life is too short smart but hard !
Probably because sizeof(int) on your system is 4.Why is this 4?
Use strlen() from <cstring> ?I would really like to know how I get the size of a char array like this
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Thanks you!, that is just what I wantedUse strlen() from <cstring> ?
life is too short smart but hard !
Test this
And see what you can come up with!Code:char first[]="123456789"; char* second = "123456789"; cout<<sizeof(first)<<endl; cout<<sizeof(second);
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.
Actually, it's because the sizeof the pointer is 4 bytes.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 = "123456789"; cout << "size of f is " << sizeof(f) << endl;Code:char *f; f = new char[20]; strcpy(f,"123456789"); cout << "size of f is " << sizeof(f) << endl; cout << f << endl; delete[] f;
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
Sure it's initialized, it's initialized to the address of the string-literal. It should however probably be a const char*.
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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.Actually, it's because the sizeof the pointer is 4 bytes.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
>>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.
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>>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.
...because pointers like f store addresses. "1234567890" is a 'string literal' and it is placed in memory somewhere, and the following assignment:char *f;
f = "123456789";
cout << "size of f is " << sizeof(f) << endl;
Why is this 4? Everytime I compile it, its 4.
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.
Last edited by 7stud; 03-11-2005 at 02:18 AM.
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
life is too short smart but hard !
