Rating:
# Challenge Name : Formatting
# Description :
Its really easy, I promise
## Analysing the binary
The decompilation from cutter is as follows
```
undefined8 main(void)
{
int32_t iVar1;
char *s;
undefined var_79h;
int64_t var_78h;
int64_t var_20h;
undefined4 var_18h;
int64_t var_14h;
int64_t var_8h;
var_14h._0_4_ = 0x66;
var_18h = 0x6c;
var_78h._0_1_ = 0;
s._0_1_ = 0x44;
s._1_1_ = 0x55;
s._2_1_ = 0x43;
s._3_1_ = 0x54;
s._4_1_ = 0x46;
s._5_1_ = (undefined)_brac0;
var_79h = (undefined)_brac1;
iVar1 = sprintf((int64_t)&s + 6, _fmt, "d1d_You_Just_ltrace_", _this, _crap, _is, _too, _easy, _what, _the, _heck);
var_20h = (int64_t)iVar1;
*(char *)((int64_t)&s + var_20h + 6) = (char)_brac1;
puts(_flag + 6);
return 0;
}
```
The decompilation pretty much shows what the flag will actually look like and what can be the result.
However, on executing the binary, the result is :
```
./formatting
haha its not that easy}
```
## Dynamic analysis of the binary
On loading the binary in gdb place a breakpoint at the puts@plt call. As soon as it hits the breakpoint there, you will get the flag in the stack.
```
gef➤ break *0x0000555555555236
```
