char c[] = "12345678"
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
Author
Quote:Original post by Boder
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
all char is number. so '\0\ is exact equal to 0. maybe I wrong.but it often confuse people. I thing \0 is exact a number it is 0.
Quote:Original post by derek7Yes, \0 equals ASCII 0, but ASCII 0 does not equal '0'. That's got a code of 48 or something (Don't know the number off the top of my head)Quote:Original post by Boder
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
all char is number. so '\0\ is exact equal to 0. maybe I wrong.but it often confuse people. I thing \0 is exact a number it is 0.
Quote:Original post by BoderActually the string 12345678 is assigned to "c", because "c" is an array. If "c" was a char pointer, you'd be correct.
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
Author
Quote:Original post by Evil SteveQuote:Original post by derek7Yes, \0 equals ASCII 0, but ASCII 0 does not equal '0'. That's got a code of 48 or something (Don't know the number off the top of my head)Quote:Original post by Boder
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
all char is number. so '\0\ is exact equal to 0. maybe I wrong.but it often confuse people. I thing \0 is exact a number it is 0.
if('\0' == 0)::MessageBox(0," equal ",0,0);
if('0' == 48)::MessageBox(0," equal ",0,0);
so \0 = number 0
Quote:Original post by Anonymous PosterQuote:Original post by BoderActually the string 12345678 is assigned to "c", because "c" is an array. If "c" was a char pointer, you'd be correct.
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
Actually, an array is a pointer. I was lied to as a child.
Quote:Original post by derek7
if('\0' == 0)::MessageBox(0," equal ",0,0);
if('0' == 48)::MessageBox(0," equal ",0,0);
so \0 = number 0
Uhh, you've just shown that the ASCII code for -
'\0' == 0
'0' == 48
Therefore '\0' != '0'.
Quote:Original post by MushuQuote:Original post by Anonymous PosterQuote:Original post by BoderActually the string 12345678 is assigned to "c", because "c" is an array. If "c" was a char pointer, you'd be correct.
The compiler creates a place in memory with "123456789\0" not "1234567890" and points c to it.
But I can say that "c" is probably not assigned 12345678, it is given a memory address.
Actually, an array is a pointer.
No.
John B
I can only conclude that deep down, all arrays have an unquenchable desire to be pointers, as they will convert themselves to pointers at the slightest provocation. It's kind of sad, really. Hopefully one day they'll be more comfortable with themselves and not try to be what they assume everyone wants them to be. They will not find happiness by pretending to be something they're not.
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