forcing a compile time error

M

Michael Gaab

How would I force the compiler to throw an error for the following:

function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.

At any rate how do I force the compiler throw a error?

thanks, Mike
 
S

Severin Ecker

function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
thats because a char(acter) is nothing more than a number (char from -128 to
127 or unsigned char from 0 to 255)
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.
there's no much difference to, say implicitly converting a short to a long
At any rate how do I force the compiler throw a error?
nope, sorry

regards,
sev
 
J

Jack Adam

Michael said:
function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.

This is normal. There is an implicit type conversion from char to short
and this is part of the C++ language, not some special compiler feature.

If you want to have different behaviors for function foo depending on
the exact type of the parameter, one solution is to to overload function
foo.
 
L

lilburne

Michael said:
How would I force the compiler to throw an error for the following:

function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.

At any rate how do I force the compiler throw a error?

The normal way to force a compilation error is to use the
#error preprocessor directive. Why you'd want to cause a
compilation error specifically on foo though is a bit of a
mystery but there are probably a myriad of ways of doing so

foo('d')
foo();
foo(&d);
 
M

Michael Kochetkov

Jack Adam said:
This is normal. There is an implicit type conversion from char to short
and this is part of the C++ language, not some special compiler feature.
I am afraid you are wrong. There is not implicit type conversion from char
to short in C++ language. There is the integral promotion from char to int
(or unsigned int) that is not an integral conversion. So, as far as char is
not an integer type, though signed char and unsigned char are integers, it
shall be promoted to int and then converted (integral convertion) to short
to make the foo(short) call.
BTW that is why void foo(int); is better then void foo(short); for foo('d')
call.
 
J

Jack Klein

thats because a char(acter) is nothing more than a number (char from -128 to
127 or unsigned char from 0 to 255)

No, a character is nothing more than a number between CHAR_MIN, which
must be at least -127 if char is signed or 0 if char is unsigned, and
CHAR_MAX which must be at least 127 if char is signed or 255 if char
is unsigned. No C++ implementation is required to support a character
with a value of -128, although some do.

And on one compiler that I use a lot these days, a char can have any
value between -32768 and +32767.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++ ftp://snurse-l.org/pub/acllc-c++/faq
 
J

Jack Adam

Michael said:
I am afraid you are wrong. There is not implicit type conversion from char
to short in C++ language. There is the integral promotion from char to int
(or unsigned int) that is not an integral conversion. So, as far as char is
not an integer type, though signed char and unsigned char are integers, it
shall be promoted to int and then converted (integral convertion) to short
to make the foo(short) call.


I know. There is one word missing from my statement, which should read:

"There is an implicit type conversion _sequence_ from char to short and
this is part of the C++ language, not some special compiler feature."

Thanks for your errata. Hope that's clear now.
 
G

Gianni Mariani

Michael said:
How would I force the compiler to throw an error for the following:

function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.

At any rate how do I force the compiler throw a error?

One technique I use is to create a template and make a
condition in the template to cause a compile time
error when the offending condition is instantiated.


void foo(short);

inline void call_foo( short s )
{
foo( s );
}

template <typename T>
inline void foo(T d)
{
struct compile_error_if_char
{
int array[ 1+ 1 / (sizeof(T)-1) ];
} x;

call_foo( d );
}


int main()
{
short x = 1;

foo( x );

foo('s');

foo( 22 );
}
 
N

NFish

Michael said:
How would I force the compiler to throw an error for the following:

function signature -
void foo(short);

function call -
foo('d');

My compiler does not complain when I call foo() with a character argument.
I am assuming that the compiler does some type casting or that char's are
interpreted as a byte. Not exactly sure.

At any rate how do I force the compiler throw a error?

thanks, Mike

~) cat test.cpp
void foo(short) {
}

#define foo(a) (1 / (sizeof(a) == sizeof(short)), (foo)(a))

int main(void) {
foo('d');
return 0;
}

~ ) c++ test.cpp
test.cpp: In function `int main()':
test.cpp:7: warning: division by zero in `1 / 0'
 

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