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Re: [CPROG] pass by reference



Det er rigtigt nok, men du kunne godt bare kalde funktionen:

void calculate_value( float *value )
  {
  *value=2.0*5.4;
  } 

I dit main-program bruger du alligevel ikke funktionens returværdi til
noget

m.v.h.
Søren


sslug@sslug wrote:
> 
> Jeg fik selv løst det ;-)
> 
>  float* calculate_value(float* value){
>    *value= 2*5.4;
>    return value;
>  }
> 
>  int main(int argc, char **argv){
>    float value;
> 
>    calculate_value(&value);
> 
>    printf("value= %f\n", value); //skriver 0.0 i stedet for 10.8
>  }
> 
> > Kære c-sprog,
> >
> > hvordan får jeg flg. eksempel til at tildele 'value' i main værdien
> > beregnet i calculate_value?
> >
> > float calculate_value(float value){
> >   value= 2*5.4;
> >   return value;
> > }
> >
> > int main(int argc, char **argv){
> >   float value;
> >
> >   calculate_value(value);
> >
> >   printf("value= %f\n", value); //skriver 0.0 i stedet for 10.8
> > }
> >
> > med venlig hilsen
> > Marie Louise


 
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