passing an argument by reference

[Michael G]

I have an api that uses reference as arguments.

void function(double z, short &x, short &y){}
some calculations are done on z and then the results are being passed back
out in x and y.

The values the i need to pass to this function are pointer to shorts.

short * x1;
short * y1;

If I call the function like

function(z1, *x1, *y1);

The compiler doesn't complain and I get correct results with the small
number of tests that I have performed on it. But I am not sure that this is
correct.

Thanks, Mike

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[Jack Klein]

On Sun, 2 May 2004 22:50:03 -0600, "Michael G" <mi****@montana .com>
wrote in comp.lang.c++:

I have an api that uses reference as arguments.

void function(double z, short &x, short &y){}
some calculations are done on z and then the results are being passed back
out in x and y.

The values the i need to pass to this function are pointer to shorts.

short * x1;
short * y1;

If I call the function like

function(z1, *x1, *y1);

The compiler doesn't complain and I get correct results with the small
number of tests that I have performed on it. But I am not sure that this is
correct.

Thanks, Mike

Assuming that x1 and y1 are properly initialized and actually point to
short ints, this is just fine.

What are passed are references to the shorts that the pointers point
to.

--
Jack Klein
Home: http://JK-Technology.Com
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http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Thomas Wintschel]

"Michael G" <mi****@montana .com> wrote in message
news:40******** **@corp.newsgro ups.com...

I have an api that uses reference as arguments.

void function(double z, short &x, short &y){}
some calculations are done on z and then the results are being passed back out in x and y.

The values the i need to pass to this function are pointer to shorts.

short * x1;
short * y1;

If I call the function like

function(z1, *x1, *y1);

The compiler doesn't complain and I get correct results with the small
number of tests that I have performed on it. But I am not sure that this is correct.

Thanks, Mike

The function call is correct provided x1 and y1 actually point to
shorts, i.e.

short a;
short b;

short * x1 = &a;
short * x2 = &b;

[E. Robert Tisdale]

Michael G wrote:

I have an api that uses reference as arguments.

void function(double z, short &x, short &y);
This is a very bad idea.
some calculations are done on z
and then the results are being passed back out in x and y.
*Real* C++ programmers do it like this:

cat main.cc

#include <iostream>

std::pair<short , short> f(double z);

int main(int argc, char* argv[]) {
std::pair<short , short> p = f(33.0);
std::cout << p.first << ", " << p.second << std::endl;
return 0;
}

[Boris Sargos]

std::pair<short , short> f(double z);

Why should we do simple when we can do complicated ?

[John Harrison]

"Boris Sargos" <bs*****@wanado o.fr> wrote in message
news:c7******** **@news-reader2.wanadoo .fr...

std::pair<short , short> f(double z);

Why should we do simple when we can do complicated ?

Which one do you consider to be more complicated?

john

[Peter Koch Larsen]

"E. Robert Tisdale" <E.************ **@jpl.nasa.gov > wrote in message news:<40******* *******@jpl.nas a.gov>...

Michael G wrote:

I have an api that uses reference as arguments.

void function(double z, short &x, short &y);

This is a very bad idea.

some calculations are done on z
and then the results are being passed back out in x and y.

*Real* C++ programmers do it like this:

> cat main.cc

#include <iostream>

std::pair<short , short> f(double z);

int main(int argc, char* argv[]) {
std::pair<short , short> p = f(33.0);
std::cout << p.first << ", " << p.second << std::endl;
return 0;
}

Well - those two functions are different. Your function has no access
to the initial values of x and y.

/Peter

[jeffc]

"Michael G" <mi****@montana .com> wrote in message
news:40******** **@corp.newsgro ups.com...

I have an api that uses reference as arguments.

void function(double z, short &x, short &y){}
some calculations are done on z and then the results are being passed back
out in x and y.

The values the i need to pass to this function are pointer to shorts.
Well, not exactly. You need to *get* the values from pointers to shorts.

short * x1;
short * y1;

If I call the function like

function(z1, *x1, *y1);

The compiler doesn't complain and I get correct results with the small
number of tests that I have performed on it. But I am not sure that this is correct.

It is. It might help to look at it like this.

short x2;
short y2;
x2 = *x1;
y2 = *y1;
function(z1, x2, y2);
Make more sense now?

[jeffc]

"E. Robert Tisdale" <E.************ **@jpl.nasa.gov > wrote in message
news:40******** ******@jpl.nasa .gov...

Michael G wrote:

I have an api that uses reference as arguments.

void function(double z, short &x, short &y);

This is a very bad idea.

some calculations are done on z
and then the results are being passed back out in x and y.

*Real* C++ programmers do it like this:

Oh boy. That's just what the OP needs......

[E. Robert Tisdale]

Peter Koch Larsen wrote:

E. Robert Tisdale wrote:

Michael G wrote:

I have an api that uses reference as arguments.

void function(double z, short &x, short &y);
This is a very bad idea.

some calculations are done on z
and then the results are being passed back out in x and y.

*Real* C++ programmers do it like this:

> cat main.cc

#include <iostream>

std::pair<short , short> f(double z);

int main(int argc, char* argv[]) {
std::pair<short , short> p = f(33.0);
std::cout << p.first << ", " << p.second << std::endl;
return 0;
}

Well - those two functions are different.
Your function has no access to the initial values of x and y.

Mike's original description of the API
describes x and y as outputs -- not inputs.
But here is a version which passes them as inputs:
cat main.cc

#include <iostream>

std::pair<short , short> f(short x, short y, double z);

int main(int argc, char* argv[]) {
short x = 7, y = 13;
std::pair<short , short> p = f(x, y, 33.0);
x = p.first;
y = p.second;
std::cout << x << ", " << y << std::endl;
return 0;
}