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#define my_sizeof(type) \
({typeof(type) x; (char *)(&x+1)-(char*)(&x);})

#define my_sizeof(type) (char *)(&type+1)-(char*)(&type)

#define my_sizeof(type) \
( (char*)(((typeof(type)*)0)+1) - (char*)((typeof(type)*)0))

#define __SIZEOF(x) \
({ \
struct { \
typeof(x) y; \
char c[0]; \
} __attribute__((__packed__)) s; \
(size_t) (&((typeof(s)*) 0)->c); \
})

// __SIZEOF(void) gives an error

#define sizeof(type) (int)(type+1) - (int)type

type *ptr
sizeof(ptr)

with the below solution #define my_sizeof(type) (char *)(&type+1)-(char*)(&type)

if i write the code like this,
int i=0;
printf("\n%d",my_sizeof(int));
compiler gives an error any idea how to resolve from this problem?

Answer given by amit5624 works.
But Can someone please elaborate it?

#define my_sizeof(type) (char *)(&type+1)-(char*)(&type)
From anonymous is correct, however, i'd suggest the use of void pointers

Size of a data type varies from architecture to archotecture. The easiest way is to declare an array of two elements and see the difference in their address. This will give the exact size of the data type !

Size of a data type varies from architecture to archotecture. The easiest way is to declare an array of two elements and see the difference in their address. This will give the exact size of the data type !

sizeof (char* type){
void *p;
map <char[ ], int> my_map ;
my_map["int"] = 1;
my_map["char"] = 2
my_map["float"] = 3;
.
.
.
switch(my_map[type]){
case 1 :
{
 p = (int*) malloc(1);
 return ((int*) &(p+1) - (int*) &p) ;
}
case 2 :
{
 p = (char*) malloc(1);
 return ((char*) &(p+1) - (char*) &p);
}
.
.
.
.
and so on
            
}

#define my_sizeof(type) \
( (char*)(((typeof(type)*)0)+1) - (char*)((typeof(type)*)0))

#define my_sizeof(type) ((type *)0+1)

This is simpler

typeof is a non standard extension of c which return the static type of the variable. So if we wish to answer this question in native C domain I suppose better way is to us size_t which is a native block data type.

#define sizeof(var)  size_t((&var)+1) - size_t(&var)

this should work.