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I have allocated dynamic memory for one integer variable
11406808Code:#include<stdio.h>#include<stdlib.h>
int main (void)
{
int *i = (int*)malloc(1 * sizeof(int));
printf(" : %d \n", i);
printf("%d \n", i, *(i));
printf(" %d \n", i, &i);
return 0;
}
11406808
11406808
if you look at the operator all they are the different * and &
Why does all three operator give a same data ?
updated : size 1
1st- You are not allocating "one" integer, but 5
2nd- You are printf the same variable (i)
3rd- You are getting warnings from your compiler and not paying attention to them.
I think *i means data store at the addressQuote:
Originally Posted by flp1969
&i means address of data
Both are different then why code gives same output ?
Take a good look at your printf() calls...Quote:
Originally Posted by abhi143
Am I missing %p to print the address ?Quote:
Originally Posted by flp1969
12325800Code:#include<stdio.h>
#include<stdlib.h>
int main (void)
{
int *i = (int*)malloc(1 * sizeof(int));
printf("%d \n", i);
printf("%d \n", i, *(i));
printf("%p \n", i, &i);
return 0;
}
12325800
00BC13A8
Well, yeah, but there is a another error... look closer... a little longer...Quote:
Originally Posted by abhi143
Take a look at thisQuote:
Originally Posted by flp1969
Code:#include<stdio.h>
#include<stdlib.h>
int main (void)
{
int *i = (int*)malloc(1 * sizeof(int));
printf("%d \n", i);
printf("%d \n", *(i));
printf("%p \n", &i);
return 0;
}
See? Very easy to spot, isn't it?Quote:
Originally Posted by abhi143
Now some considerations:
Compiling and running:Code:#include <stdio.h>
#include <stdlib.h>
int main( void )
{
// You don't need to use type casting here, malloc() will return a "generic" pointer to void, always.
int *i = malloc( sizeof( int ) );
printf( "%p\n", i ); // Use %p to specify the argument is a pointer.
printf( "%d\n", *i ); // Will print the integer pointed by i, but you know it is uninitialized, don't you?
printf( "%p\n", &i ); // This will make the compiler allocate 'i' on stack (so you can get the "adddress of" i.
// You'll get the address of 'i', not the address of the dynamic block allocated.
// Not really necessary, but it is a good pratice to do a little housekeeping.
free( i );
}
It is only a coincidence *i is zero, could be anything...Code:$ cc -o test test.c
$ ./test
0x55a7d5516260
0
0x7ffc8f1c66a0
we have allocated memory space for variable iQuote:
Originally Posted by flp1969
If I want to assign the value to variable What I have to do
You could say that, but it would be more accurate to say that you allocated memory for what i points to.Quote:
Originally Posted by abhi143
Just remember that since i is a pointer to int, *i is an int, so you assign to the int variable as you would any other int variable:Quote:
Originally Posted by abhi143
Code:*i = 123;
correctQuote:
Originally Posted by laserlight
outputCode:#include<stdio.h> #include<stdlib.h>
int main (void)
{
int *i = (int*)malloc(1 * sizeof(int));
*i = 123;
printf("%d \n", i);
printf("%d \n", *(i));
printf("%p \n", &i);
return 0;
}
10949544
123
0061FF2C
Now If I want to assign value to array
I can't writeHow to assign array valueCode:array[5] = {1,2,3,4,5}; //static array
Code:#include<stdio.h>#include<stdlib.h>
int main (void)
{
int size = 5, i=0;
int *array= (int*)malloc((sizeof(int)) * size);
for(i=0; i < size; i++)
{
}
return 0;
}
Use the operador [] or the operator *:Quote:
Originally Posted by abhi143
Code:int *p;
p = malloc( size * sizeof( int ) );
for ( i = 0; i < size; i++ )
p[i] = K; // OR *(p + i) = k;
No let's suppose we have allocated 5 memory location using mallocQuote:
Originally Posted by flp1969
we want to assign following values to each memory location
memory location 0 = 5
memory location 1 = 15
memory location 2 = 5
memory location 3 = 53
memory location 4 = 101
How can we do that ?
The same way you'd initialise a normal array.Quote:
No let's suppose we have allocated 5 memory location using malloc
we want to assign following values to each memory location
memory location 0 = 5
memory location 1 = 15
memory location 2 = 5
memory location 3 = 53
memory location 4 = 101
How can we do that ?
I'm not sure if you can do this with malloc but by using the new keyword in C++, you can provide an initialiser list.Code:p [0] = 5;
p [1] = 15;
p [2] = 5;
p [3] = 53;
p [4] = 101;
list initialization (since C++11) - cppreference.comCode:int* p = new int [5] { 5 , 15 , 5 , 53 , 101 };
As your doubt is posted in the C Programming section and new is part of C++, you probably wouldn't wanna be doing that and so you may do a normal index by index initialisation. If you were to take input from the user for initialization, you could:
Code:for (int i = 0; i < Size; i++)
{
// Take input at the i'th index the way it's done in C
}
If you want to initialize a bunch of elements of dynamically allocated array you can do as @Zeus_ told you:
Code:p[0] = 5;
p[1] = 15;
p[2] = 5;
p[3] = 53;
p[4] = 101;
Or, in C (not C++), with a compiler capable of compiling code compliant to ISO 9989:2011 standard (I am not sure, right now, but I believe ISO 9989:1999 already had compound literals), you can use compound literals, as in:
Notice is easier to just simply do:Code:int *p = malloc( 5 * sizeof( int ) );
memcpy( p, (int []){ 5, 15, 5, 53, 101 }, 5 * sizeof( int ) );
But you are asking how to inicialize a dynamic allocated buffer!Code:int a[] = { 5, 15, 5, 53, 101 };