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Answer 1

A:

At the start of the code place: typedef long int32; The type int32 doesn't exist sinse int already stands for "integer with 32 bits". Actually int may vary depending on the compiler, but you still can explicitely use "a 32 bits integer" using the type long.

Havenard 2009-08-29 05:37:25

shouldn't that be the other way around? i.e., "typedef int int32;"? I tried it this way and it worked.

blackkettle 2009-08-29 05:49:42

Sorry, that's completely wrong. long is generally 64 bits on a 64-bit platform. int is allowed to be a 16-bit type. If you want a specific width you should use the typedefs in <stdint.h> as caf suggests in his answer.

Jim Lewis 2009-08-29 06:28:22

I'm not familiar to 64-bit architectures, if int is allowed to be 16-bit and long is allowed to be 64-bit, howtf I explicitely declare a 32-bit integer? As far as I know, to declare a "64 bit" integer you use **long long**, not long (alone).

Havenard 2009-08-29 16:59:15

Actually, it's even more complicated. On some 64-bit systems (windows), int and long are 32 bits and long long is 64 bits (this is often called "LLP64"). On some other systems (OS X, Linux), int is 32 bits, but long and long long are both 64 bits (called "LP64"). On still other systems, int, long, and long long are all 64 bits (called "ILP64").If you want to have a (relatively) portable integer type that you know is exactly 64 bits, you should use the types int64_t and uint64_t defined in <stdint.h>.

Stephen Canon 2009-08-29 17:19:45

From Stroustrup C++ Special Edition, p.75: This is what is guaranteed about sizes of fundamental types: 1 = sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long). Also, 1 <= sizeof(bool) <= sizeof(long). Also, sizeof(char) <= sizeof(wchar_t) <= sizeof(long). Also, sizeof(float) <= sizeof(double) <= sizeof(long double). Also, sizeof(N) = sizeof(signed N) = sizeof(unsigned N). Char is at least 8 bits, short at least 16 bits, long at least 32 bits. Entirely possible for int to be 16 bits and long to be 128 bits, with no 32-bit data type.

ChrisInEdmonton 2009-11-04 18:57:09

Not only that, but it's also entirely possible for `char` to be 10 bits, and `int` to be 30 bits. On the other hand, I've been told that on some DSPs you have `sizeof(char)==sizeof(short)==sizeof(int)==2`, or 16 bits, because it can only work with words and not individual bytes.

Pavel Minaev 2009-11-05 02:38:50

Answer 2

+9 A:

The int32 type isn't standard C - the standard equivalent is to #include <stdint.h> and use int32_t.

However, as a POSIX system, on Ubuntu plain int is 32 bit so you could just use that.

caf 2009-08-29 05:38:11

Two things of note: 1) `<stdint.h>` is standard, but it's C99 - not a problem for gcc, but can be elsewhere (e.g. MSVC); and 2) POSIX doesn't require `int` to be 32-bit, and there are Unixes out there for which it's not true.

Pavel Minaev 2009-11-05 02:36:36

There are four programming environments allowed by POSIX to be the default (listed here: http://opengroup.org/onlinepubs/009695399/utilities/c99.html#tagtcjh_11 ), all of which require `int` to be at least 32 bits.

caf 2009-11-05 03:45:23

Answer 3

A:

Size of 'int' depends on the compiler you use. For same target architecture different compilers can have different sizes for 'int'. It depends on how they are utilizing the target features. For example, Turbo C running on windows has 'int' size as 2 bytes (its a DOS executable). While Visual C gives the size of 'int' as 32 (win32 executable).

In general, size of 'int' is determined by the size of word as per the target architecture. Size of word is reflected by 1. Processing size of arithmetic operations. 2. Size of general purpose registers. 3. Size of the address used to designate a location.

However in your case you are wrongly using a keyword 'int32' for a data type. Check the target machine and fix up the keyword used accordingly.

Ganesh Gopalasubramanian 2009-11-05 02:30:19

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