help on good old C

[pavansavoy]

since no one is online elsewhere, here it goes explain int *p; p=(int (*)[4])malloc (3*sixeof(*p)) please

[David Crow]

What exactly are you wanting to know about the two statements?

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"When I was born I was so surprised that I didn't talk for a year and a half." - Gracie Allen

[Micie]

int *p; // prepare int pointer for var p p=(int (*)[4])malloc (3*sizeof(*p)); This code is allocating an array for 12 bytes long (3 ints = 12bytes) in memory and returns result to variable p :-O **__________ I'm made in C++... and I'm proud of it!_**

[4apai]

this code trully allocate array of three int elements. int *p; p=(int *)malloc (3*sizeof(int));

[Henry miller]

I can't. I'm guessing sixeof is a typo and you ment sizeof. (a forgivable mistake unless you really ment sixeof, in which case I'd need to know more about it) the int (*)[4] part is wrong. It is tyring to tell the compiler that you have an array of 4 ints, but just allocated 3! Not that the compiler would check, it is legal C to then use p[5], though it is undefined behavior and will randomly crash your program. Just write int *p = malloc(3*sizeof(int)); like the rest of us would.

[Rick York]

Have a look at this : http://www.codeproject.com/cpp/complex_declarations.asp[^] __________________________________________ a two cent stamp short of going postal.