Martin Husemann <mar...@duskware.de> wrote:
> On Tue, Dec 06, 2022 at 12:34:02PM -0500, Jan Schaumann wrote:
> > #define NUM 6
> > char chararray[NUM];
> > (void)strncpy(chararray, "abc", NUM);
>
> That makes the code worse, often the defintion of the buffer and the
> strncpy are not next to antother and later changing the size of the
> buffer often leads to wrong limits for the strncpy call.
Fair enough.
How do you feel about adding a separate manual page
for the sizeof operator? (See attached proposal,
though my mandoc is not very good and likely needs
clean up.)
-Jan
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.Dd December 10, 2022
.Dt sizeof 7
.Os
.Sh NAME
.Nm sizeof
operator
.Nd yield the storage size of the given operand
.Sh SYNTAX
.Nm Vt ( type )
.br
.Nm Vt expression
.Sh DESCRIPTION
The size of primitive data types in C may differ
across hardware platforms and implementations.
It may be necessary or useful for a program to be able
to determine at runtime the storage size of a data
type or object.
.Pp
The unary
.Nm
operator yields the storage size of an expression or
data type in
.Em char sized units .
As a result, 'sizeof(char)' is always guaranteed to be 1.
(The number of bits per
.Dv char
is given by the
.Vt CHAR_BITS
definition in the
.Dv limits.h
header.)
.Sh EXAMPLES
The following example illustrate the possible results
of calling
.Nm
on an LP64 system:
.Pp
When applied to a simple variable or data type,
.Nm
returns the storage size of the data type of the
object:
.Bl -column -offset indent \
".Li sizeof(struct flex) " ".Sy Result"
.It Sy Object or type \
Ta Sy Result
.It Li sizeof(char) \
Ta 1
.It Li sizeof(int) \
Ta 4
.It Li sizeof(float) \
Ta 4
.It Li sizeof(double) \
Ta 8
.It Li sizeof(char *) \
Ta 8
.El
.Pp
For initialized data or uninitialized arrays of a
fixed size known at compile time,
.Nm
will return the correct storage size:
.Bd -literal -offset indent
#define DATA "1234567890"
char buf1[] = "abc";
char buf2[1024];
char buf3[1024] = { 'a', 'b', 'c' };
.Ed
.Bl -column -offset indent \
".Li sizeof(struct flex) " ".Sy Result"
.It Sy Object or type \
Ta Sy Result
.It Li sizeof(DATA) \
Ta 11
.It Li sizeof(buf1) \
Ta 4
.It Li sizeof(buf2) \
Ta 1024
.It Li sizeof(buf23 \
Ta 1024
.El
.Pp
When applied to a struct or union,
.Nm
returns the total number of units in the object,
including any internal or trailing padding used to
align the object in memory.
This result may thus be larger than if the storage
size of each individual member had been added:
.Bd -literal -offset indent
struct s1 {
char c;
}
struct s2 {
char *s;
int i;
}
struct s3 {
char *s;
int i;
int j;
}
struct s4 {
struct s1 a;
struct s2 b;
struct s3 c;
}
.Ed
.Bl -column -offset indent \
".Li sizeof(struct flex) " ".Sy Result"
.It Sy Object or type \
Ta Sy Result
.It Li sizeof(struct s1) \
Ta 1
.It Li sizeof(struct s2) \
Ta 16
.It Li sizeof(struct s3) \
Ta 16
.It Li sizeof(struct s4) \
Ta 40
.El
.Pp
When applied to a struct containing a flexible array
member,
.Nm
returns the size of the struct
.Em without
the array, although again possibly including any
padding the compiler deemed appropriate:
.Bd -literal -offset indent
struct flex {
char c;
char b;
char array[];
}
.Ed
.Bl -column -offset indent \
".Li sizeof(struct flex) " ".Sy Result"
.It Sy Object or type \
Ta Sy Result
.It Li sizeof(struct flex) \
Ta 2
.El
.Pp
One of the more common uses of the
.Nm
operator is to determine the correct amount of memory
to allocate:
.Bd -literal -offset indent
int *nums = calloc(512, sizeof(int));
.Ed
.Pp
The
.Nm
operator can be used to calculate the number of
elements in an array by dividing the size of the array
by the size of one of its elements:
.Bd -literal -offset indent
int nums[] = { 1, 2, 3, 4, 5 };
const int howmany = sizeof(nums) / sizeof(nums[0]);
.Ed
.Sh RESULT
The result of the
.Nm
operator is an unsigned integer type, defined in the
.Dv stddef.h
header as a
.Vt size_t .
.Sh NOTES
It is a common mistake to apply
.Nm
to a dynamically allocated array:
.Bd -literal -offset indent
char *buf;
if ((buf = malloc(BUFSIZ)) == NULL) {
perror("malloc");
}
/* Warning: wrong! */
(void)strncat(buf, input, sizeof(buf) - 1);
.Ed
.Pp
In that case, the operator will return the storage
size of the pointer ('sizeof(char *)'), not the
allocated memory!
.Pp
.Nm
determines the
.Ev size
of the result of the expression given, but does
.Em not evaluate
the expression:
.Bd -literal -offset indent
int a = 42;
printf("%ld - %d\\n", sizeof(a = 10), a); /* Result: "4 - 42" */
.Ed
.Pp
Since it is evaluated by the compiler and not the
preprocessor, the
.Nm
operator cannot be used in a preprocessor expression.
.Sh HISTORY
This manual page first appeared in
.Nx 10.0 .
.Sh AUTHORS
This manual page was written by
.An Jan Schaumann Aq Mt jscha...@netbsd.org .
