TASK_SIZE is a kernel constant that defines the upper limit of the accessible memory for the code working at the lowest privilege level.
Its value is usually set to 0xc0000000 on systems with less than 1GB of physical memory (all examples included in this article refer to this value). The memory above this limit contains the kernel code .
Is there a way to determine the running kernels TASK_SIZE through c program ??
After a lot of google search and analysis , i got a logic
Assume net virtual address are 4gb and the it is divided in 1:3 ratio.
Rough assumptions:-
Kernel (upper 1 gb): c0000000 -ffffffff
USer space (below 3gb):0-c0000000
than
#define GB *1073741824
unsigned int num;
unsigned int task_size;
task_size=(unsigned)&number+ 1 GB / 1 GB * 1GB;
[the process's stack area will be allocated below the kernel space]
so address of num (in stack)= somewhere around in 3 GB range ex:[3214369612]
now adding 1 GB = 1073741824+3214369612=4288111436
dividing by 1GB=3.993614983 that will be 3 (unsigned int)
now multiplying by 1GB = 3 *1073741824 = 3221225472 i.e. (0xC0000000 in hex)
hence i got the kernel starting address (TASK_SIZE)
I tried it assuming (2:6) ratio also and got correct result.
Is this a fair logic ,Please comment ???