Don't assume (as we've pointed out above).YeXo wrote: From what you say I assume gcc (and any windows port of that) is using C99, is that correct?
No. GCC defaults to C90 plus GNU extensions when compiling C, and to C++98 with GNU extensions when compiling C++. Note that GNU is not fully compliant to C99 even when told to use C99. It is not fully compliant to the C++ standard, either, but very few compilers are. (And yes, I have to mention it, virtually all of them are lacking [tt]export[/tt].YeXo wrote: From what you say I assume gcc (and any windows port of that) is using C99, is that correct?
)kicks self I entirely missed the "C89/C99?" part.Solar wrote: ...but in C89 there is no such thing as a line comment, so it would be (42 + 12 / 3 + 4) = 50...
Don't think of them as "adding numbers to pointers", consider them as "indexing an array".YeXo wrote: Adding numbers to pointers always confuses me.
but ptr doesn't point to "value 0x14". it points towards "address 0x14" (which is very unlikely to be a valid address anywayptr points to the value: 0x14
)Code: Select all
int i = 5;
printf("%d %d %d", i++, i++, i++);
When computer "ptr + 4" the compiler adds 4*sizeof(ptr) rather than 4*sizeof(*ptr). On a pentium, the first would mean 4*4 (the size of a pointer is 32 bits, or 4 bytes), while the later whould be 4*whatever the size of the type is.Pype.Clicker wrote: [tt]*(ptr+4)[/tt] is semantically identical to [tt]ptr[4][/tt], whatever type [tt]ptr[/tt] might have. That means when computing "ptr + 4" your compiler actually produce an that is 4*sizeof(*ptr) bytes farther than the address of ptr.
Thus on a machine where "long int" is 4 bytes, "long_int_ptr++" do advance the pointer to the next long int ...
It wasn't meant to be output of any code, merely as explanation of the contents of the variables. You're right ptr points to address 0x14, instead of value 0x14, that was my mistake. (ofcourse 0x14 is unlikey to be a valid address, but it is most times more simple to explain something with small than with large numbers).and, btw, i don't know what code has output the linebut ptr doesn't point to "value 0x14". it points towards "address 0x14" (which is very unlikely to be a valid address anywayptr points to the value: 0x14
According to the spec, the behaviour for more than one i in a line, one of more containing ++ or -- is undefined. Undefined means as much as the compiler might eat your cat and sell the fur on ebay.viral wrote: hello...
Try thisit will print:Code: Select all
int i = 5; printf("%d %d %d", i++, i++, i++);
7 6 5
Acctually arguements are passed on stack.. hence value of first i++ (5) is pushed 1st then (6) then (7) and finnally they are poped and print in as 7 6 5.
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i++++;
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(i++)++;
This first inits x to 5 in the first line. then in the if statement x is assigned 6(only one = sign) , and because true in c/c++ is non zero than it will always be true and always print "six".Pype.Clicker wrote: if _that_ confuses you, then what would it be withCode: Select all
int x = 5; if (x=6) printf("six"); else printf("not six");
The c preprocessor does not evalate the expression in a define, it just adds the string where ever it is(i.e. SIX * NINE = 1 + 5 * 8 + 1)or evenwhich does print ... 42Code: Select all
#include <stdio.h> #define SIX 1 + 5 #define NINE 8 + 1 int main(void) { printf( "What do you get if you multiply %d by %d? %d\n", SIX, NINE, SIX * NINE ); return 0; }
ahem.. what???When computer "ptr + 4" the compiler adds 4*sizeof(ptr) rather than 4*sizeof(*ptr). On a pentium, the first would mean 4*4 (the size of a pointer is 32 bits, or 4 bytes), while the later whould be 4*whatever the size of the type is.
how did you get that idea?Code: Select all
$ ./foo
sizeof(Foo)=32
&foo[4]=bfc3eaf4
foo+4 =bfc3eaf4
&ptr[4]=bfc3eaf4
ptr+4 =bfc3eaf4
wrong =bfc3ea84
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int loadOfInts[20];
// initialised here
int *ptr = &(loadOfInts[0]); // or simply int *ptr = loadOfInts;
ptr++; // ptr points to the second element
ptr = ptr + 4; // ptr points to 6th elementAre you sure you wanna learn C ??LongHorn wrote: Somebody is coming to our college to teach us C in deep. I don't know how deep is it going to be? Just like to impress them by saying c99 kind of stuff. Is there any book explaining c89,c99 compliance etc.
What I meant is this:zloba wrote: 2 YeXo:ahem.. what???When computer "ptr + 4" the compiler adds 4*sizeof(ptr) rather than 4*sizeof(*ptr). On a pentium, the first would mean 4*4 (the size of a pointer is 32 bits, or 4 bytes), while the later whould be 4*whatever the size of the type is.