memory allocation for struct in c++

amused · 03-30-2010, 12:14 AM

How do we allocate memory of struct?

what i did was

Code:

int main()
struct amp
{
    string z;
    char y;
    float x;
    int w;
};

cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"and it is located at"<<struct amp*s = malloc(sizeof(struct amp))<<endl;

it gives me an error---

In funtion 'int main()':
error: expected primary-expression before 'struct'
error: expected ';' before 'struct'

Please help me with this.. i tried looking it on net but could not find it..

amused · 03-30-2010, 12:20 AM

How do we allocate memory of struct?

what i did was

Code:

int main()
struct amp
{
    string z;
    char y;
    float x;
    int w;
};

cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"and it is located at"<<struct amp*s = malloc(sizeof(struct amp))<<endl;

it gives me an error---

In funtion 'int main()':
error: expected primary-expression before 'struct'
error: expected ';' before 'struct'

Please help me with this.. i tried looking it on net but could not find it..

paulsm4 · 03-30-2010, 12:26 AM

Why would you do something perverse like mix cout (C++) with "malloc()"

?

Anyway, the problem is mixing a declaration ("struct amp*s) in with an expression.

You have several alternatives:

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
} s;

int
main(int argc, char *argv[])
{
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from stack ("automatic" variable)
  struct amp s;
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from heap 
  // "new" is a C++ operator, "malloc()" is a C/C++ standard library function
  struct amp *s = new struct amp;
  cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"\n";

amused · 03-30-2010, 12:31 AM

Thanks for the code.

but my problem was on memory location, that is my second part struct amp*s = malloc(sizeof(struct amp)). please help me with this.

Thanks again.

Quote:

Originally Posted by paulsm4

Why would you do something perverse like mix cout (C++) with "malloc()"

?

Anyway, the problem is mixing a declaration ("struct amp*s) in with an expression.

You have several alternatives:

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
} s;

int
main(int argc, char *argv[])
{
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from stack ("automatic" variable)
  struct amp s;
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from heap 
  // "new" is a C++ operator, "malloc()" is a C/C++ standard library function
  struct amp *s = new struct amp;
  cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"\n";

paulsm4 · 03-30-2010, 12:33 AM

Hi -

Just like deja vu all over again

Anyway, the problem is mixing a declaration ("struct amp*s) in with an expression.

You have several alternatives:

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
} s;

int
main(int argc, char *argv[])
{
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from stack ("automatic" variable)
  struct amp s;
  cout <<"The size of 'struct' is"<< sizeof(s)<<"\n";

Code:

struct amp
{
    string z;
    char y;
    float x;
    int w;
};

int
main(int argc, char *argv[])
{
  // Allocate from heap 
  // "new" is a C++ operator, "malloc()" is a C/C++ standard library function
  struct amp *s = new struct amp;
  cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"\n";

pixellany · 03-30-2010, 12:40 AM

I have merged your 2 duplicate threads---next time, only one thread per topic---thanks

amused · 03-30-2010, 12:52 AM

i am not getting it.. i just wanna know where is struct located.. please help me with finding how to find the memory location OF struct.

graemef · 03-30-2010, 03:11 AM

Quote:

Originally Posted by amused

i am not getting it.. i just wanna know where is struct located.. please help me with finding how to find the memory location OF struct.

If your variable is a pointer then it's value is the address location.
Otherwise you can find the address by using the & operator.

Why do you want to find the memory location of struct?

paulsm4 · 03-30-2010, 12:00 PM

Hi -

Graemef is absolutely correct:

Quote:

If your variable is a pointer then it's value is the address location.
Otherwise you can find the address by using the & operator.

For example:

Code:

#include <stdio.h>

#define MAX_STRING 80

struct amp
{
    char z[MAX_STRING+1];
    char y;
    float x;
    int w;
};

int 
main(int argc, char *argv[])
{
  struct amp my_struct, *my_struct_p;

  my_struct_p= &my_struct;
  printf ("addr(my_struct)= 0x%x == 0x%x...\n", 
    &my_struct, my_struct_p);
  return 0;
}

'Hope that helps .. PSM

smeezekitty · 03-30-2010, 12:04 PM

You cannot declare there.

Quote:

Originally Posted by amused

How do we allocate memory of struct?

what i did was

Code:

int main()
struct amp //<-- WTF?
{
    string z;
    char y;
    float x;
    int w;
};

cout <<"The size of 'struct' is"<< sizeof(struct amp)<<"and it is located at"<<struct amp*s = malloc(sizeof(struct amp))<<endl;

it gives me an error---

In funtion 'int main()':
error: expected primary-expression before 'struct'
error: expected ';' before 'struct'

Please help me with this.. i tried looking it on net but could not find it..