2 ms·

That depends how much the compiler can optimize (away). If the next call is free(), it's quite trivial to skip the zeroing and just take the correct branch. I

• zAy0LfpBZLC8mAC 12y ago
  Why shouldn't the compiler be able to figure out that you sum a series of numbers that aren't used anywhere else, thus don't need to be spilled to RAM?
  • xxs 12y ago
    You write them to a global state volatile
    • zAy0LfpBZLC8mAC 12y ago
      How does that require writing the random data to RAM, apart from the volatile variable itself?
• gizmo686 12y ago
  You still are not guaranteed to clear the buffer like that. for (int i=0; i<len; i++){ sensitiveBuffer[i]=random(); } int sum=0; for (int i=0; i<len; i++){ sum+=sensitiveBuffer[i]; } volitileVar=sum; Using loop fusion, the compiler can optimize this to: int sum=0; for (int i=0; i<len; i++){ sensitiveBuffer[i]=random(); sum+=sensitiveBuffer[i]; } volitileVar=sum; Which it can then optimize to: int sum=0; for (int i=0; i<len; i++){ sum+=random(); } volitileVar=sum; In fact, as the article points out, the compiler can legally transform: reallyZeroBuffer(sensitiveBuffer); into pointlesslyCopy(sensitiveBuffer); reallyZeroBuffer(sensitiveBuffer);
  • xxs 12y ago
    not like that: Not exactly like that: I didn't mean simple that way, more like hash alike. The 1st example is not optimal but shows the idea. I disagree about "pointlessCopy". Of course it's permitted by the standard but it's not an optimization. Using such a broken compiler is beyond help. ------- volatileVar *= random(); for (int i=0;i<len;i++){ volatileVar+=buf[random()%len]+0x61c88647; buf[i]=volatileVar; } ===== static volatile uint volitileVar; for (int i=0; i<len; i++){ sensitiveBuffer[i]=random(); } for (int i=0; i<len; i++){ volitileVar+=sensitiveBuffer[volatileVar%len]; }