Smashing the stack example3.c confusion
Article can be found here . I'm reading up on smashing the stack and have found myself to be getting stuck on example3.c. 0x80004a3 <main+19>: call 0x8000470 <function> 0x80004a8 <main+24>: addl $0xc,%esp 0x80004ab <main+27>: movl $0x1,0xfffffffc(%ebp) 0x80004b2 <main+34>: movl 0xfffffffc(%ebp),%eax The author indicates that we want to skip from 0x80004a8 to 0x80004b2 and that this jump is 8 bytes; how has the author determined this is 8 bytes? I have recreated the code and sent it through objdump and found that it's not 8 bytes (I am on a 64 bit machine but I've made sure to compile using 32