Odtnhj

Hi I'm trying to create a function that takes a void* pointer as parameter and print all element of the vector using printf

the declaration of the function is:

void print_vec(void *vec,int dime_se,int dime,char *format);

and implementation is:

void print_vec(void *vec,int dime_se,int dime,char *format)
 for(int i=0;i<dime;i++)
 printf(format,*(vec+dime_se*i));

the problem is that when i compile the compiler returns:

error: invalid use of void expression printf(format,*(vec+dime_se*i));

So the question there is a way to do this task without make this?

void print_vec(void *vec,int dime_se,int dime,char *format)
for(int i=0;i<dime;i++)
 switch(format[1])
 case 'c':
 printf(format,*((char*)(vec+dime_se*i)));
 break;

 case 'd':
 printf(format,*((unsigned int*)(vec+dime_se*i)));
 break;


 case 's':
 printf(format,*((char**)(vec+dime_se*i)));
 break;

 case 'i':
 printf(format,*((int*)(vec+dime_se*i)));
 break;



 case 'f':
 printf(format,*((float*)(vec+dime_se*i)));
 break;

 default:
 break;
 

You don't need the dime_se param. The size is more or less implicit in the format character.
Also, if you cast to the right pointer type you can let C do the scaling instead of doing it
manually:

#include <stddef.h>
#include <stdio.h>
void print_vec(void const*vec,size_t dime,char const*format)

 for(size_t i=0;i<dime;i++)
 switch(format[1])
 #define X(Char,Tp) break;case Char: printf(format,((Tp*)vec)[i])
 X('c',char);
 X('d',unsigned);
 X('s',char*);
 X('i',int);
 X('f',float);
 break;default: return;
 #undef X
 

I said more or less because type promotion kind of blurs the line. %f could well mean you've got a double instead of a float.

If you want to use a dime_se parameter to disambiguate that case, you can:

 break;case 'f': 
 if(sizeof(double)==dime_se)
 printf(format,((double*)vec)[i]);
 else 
 assert(sizeof(float)==dime_se); 
 printf(format,((float*)vec)[i]); 
 

but I'm not sure if a print_vec function like that is a good idea.