Timeline for What is the output for this and how?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2023 at 3:20 | answer | added | user207558 | timeline score: -1 | |
| Mar 5, 2023 at 1:37 | review | Close votes | |||
| Mar 19, 2023 at 3:03 | |||||
| Mar 4, 2023 at 14:53 | comment | added | Satyam Singh | Ok Thanks Got it. | |
| Mar 4, 2023 at 11:58 | comment | added | blabb | Yes correct overflow will be discarded byte is 8 bit max 0xff min 0x00. Ox14d is word or 16 bit. So obviously 0x100 is discarded | |
| Mar 4, 2023 at 10:45 | comment | added | Satyam Singh | I am still little confused.... as this is entire function {String_from_char_((char *)&Decoded_string, *(_BYTE *)(Addressforencoded + Integer_count - 1) + 79);} now as you said {*(_BYTE *)(Addressforencoded + Integer_count - 1)} --> 0xFE and then addition happen so 0xFE+0x4f(hexof 79) --> gives 0x14D but only 4D has been written in memory | |
| Mar 4, 2023 at 4:45 | comment | added | blabb | Key is a pointer it adds 79 to each byte upto integer count. Assume if key to key + intcount contains 0 then after this loop key to key+int count will all be 79 memset (* , 79 , intcount); | |
| Mar 3, 2023 at 20:54 | history | edited | Satyam Singh | CC BY-SA 4.0 |
deleted 25 characters in body
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| Mar 3, 2023 at 20:53 | history | edited | Satyam Singh | CC BY-SA 4.0 |
added 4 characters in body
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| S Mar 3, 2023 at 20:51 | review | First questions | |||
| Mar 5, 2023 at 1:21 | |||||
| S Mar 3, 2023 at 20:51 | history | asked | Satyam Singh | CC BY-SA 4.0 |