Open nixawk opened 7 years ago
Notice: please use x86 platform or -m32.
#!/bin/bash
for file in `ls *.c`;do gcc $file -o "${file/\.c/\.out}";done
ls *.out | xargs size
text data bss dec hex filename
1484 300 4 1788 6fc main.out
1484 304 4 1792 700 global_var_init.out
1484 304 4 1792 700 static_var_init.out
1484 300 8 1792 700 global_var_uninit.out
1484 300 8 1792 700 static_var_uninit.out
Processes running under the user space have access only to a limited part of memory, whereas the kernel has access to all of the memory. Processes running in user space also don't have access to the kernel space. User space processes can only access a small part of the kernel via an interface exposed by the kernel - the system calls. If a process performs a system call, a software interrupt is sent to the kernel, which then dispatches the appropriate interrupt handler and continues its work after the handler has finished.
Kernel space code has the property to run in "kernel mode", which (in your typical desktop -x86- computer) is what you call code that executes under ring 0. Typically in x86 architecture, there are 4 rings of protection. Ring 0 (kernel mode), Ring 1 (may be used by virtual machine hypervisors or drivers), Ring 2 (may be used by drivers, I am not so sure about that though). Ring 3 is what typical applications run under. It is the least privileged ring, and applications running on it have access to a subset of the processor's instructions. Ring 0 (kernel space) is the most privileged ring, and has access to all of the machine's instructions. For example to this, a "plain" application (like a browser) can not use x86 assembly instructions lgdt to load the global descriptor table or hlt to halt a processor.
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