With the front: + + (-) There are too many confusing places, (i++) + (i++) and (++i) + (++i) What is the difference? If you understand it from the machine's point of view, it will be enlightened.
Let's take a look at the procedure:
int main() { int i=3; int j=(i++)+(i++); // int j=(++i)+(++i); printf("%d,%d\n",i,j); }
(1) Under VC 6.0:
for (i++) + (i++):
Result: i=5,j=6
The corresponding assembly code is (with detailed comments):
8B 45 FC mov eax,dw
Author: serious snowAfter a user clicks log on to the game, the server sends a piece of data to the local device: The selected part is the random key sent from the server to the Local Machine (it is not known that the key combination is inappropriate because it will be encrypted as data ).... The rest are some data packet features and offset sizes...Then this key is used for a series of processing .... The following authentication information is sent to the server: The data in the selected part
Reference: http://blog.csdn.net/eagler_hzh/article/details/6550841
In fact, the function to be extracted is the void Xid _cpu_detect (void) in x264 \ common \ CPU. C. The source file
int x264_cpu_cpuid_test( void );void x264_cpu_cpuid( uint32_t op, uint32_t *eax, uint32_t *ebx, uint32_t *ecx, uint32_t *edx );void x264_cpu_xgetbv( uint32_t op, uint32_t *eax, uint32_t *edx );uint32_t x264_cpu_detect( void ){
;}}Else{Float F = (float) SQRT (x * x + y * Y + z * z );If (F! = 0.0f){F = 1.0f/F;X * = f; y * = f; z * = F;}}}
// Calculate the cross multiplication using two vectors and save the result to this vector.Void vector: Cross (const vector * pu, const vector * PV){If (g_busesse2){_ ASM{MoV eax, Pu;MoV edX, PV;
Movups xmm0, [eax]Movups xmm1, [edX]Movaps xmm2, xmm0Movaps xmm3, xmm1
Shufps xmm0, xmm0, 0xc9Shufps x
have been standing behind the scenes, and some things all the ins and outs only I know, because I and Dr. Huanghai, NetEase Cloud class, Professor Wunda and Coursera GTC translation platform, Deeplearning.ai official have had exchanges, so I still have to leave something as a description, Save everyone in the network every day noisy ah did not calm down to study seriously. As mentioned in this article, I have a chat record to support, some of the auth
functions are slightly different from the previous ones. The print function appears virtual before. However, this virtual has played a huge role. It is no exaggeration to say that, without virtual functions, there is basically no design pattern, which cannot reflect the great superiority of the C ++ language in object-oriented design. Let's take a look at how this virtual works?
76: employee p;
0040128D lea ecx, [ebp-10h]
00401290 call @ ILT + 45 (employee: employee) (00401032)
00401295 mov d
, for a loop statement, the%EBX value is increased by 1, and when%EBX is no more than 5 o'clock, repeat the process, i.e.%ebx=%ebx+1;%eax=%ebx+1,%eax=%eax* the value of the previous validated number, comparing the%EAX to the value currently being validatedTherefore the first value is 1, the second value should be (+) *1=2, the third value is (2+1) *2=6, the fourth value is (3+1) *6=24, the fifth value is (4+1) *24=120, and the sixth value is (5+1) *120=720.
Phase_3
Phase_3 also cal
And the list of awesome-* series
Bayandin/awesome-awesomeness GitHub
sqlmap!
goagent!
Yes, there's shadowsocks!.
Open EdX
Open EdX is committed to creating a powerful and flexible, open and large-scale online classroom platform. Also used to study learning and distance education
After graduation, the most familiar thing is this project.
Tall may not be, technically not stunning, after all, is busine
:00440f2c |. 8b45 FC MOV eax,dword PTR ss:[ebp-4]00440f2f |.BA 14104400 MOV edx,crackme3.00441014; ASCII "Registered User"00440f34 |.E8 F32BFCFF call crackme3.00403b2c; The key is to go with F7.00440f39 |.JNZ short crackme3.00440f8c; This is the end of the jump.00440f3b |. 8d55 FC LEA edx,dword PTR ss:[ebp-4]00440f3e |. 8b83 C8020000 MOV eax,dword PTR Ds:[ebx+2c8]00440f44 |. E8 D7FEFDFF Call Crackme3.00420e
program writer's quality decision.The inline assembly is passed in C + +The results of the actual discovery of 500W data are as follows:Algorithm name inline assembly algorithm time C + + algorithm timeBubble sort 5W Data slow dying 5W data slow to deathQuick sort 600ms about 500ms around------------------Why there is a fast sorting algorithm, the results of the assembly is not a C + + efficiency is high, because I write the inline assembly is not automatically generated by the compiler high ef
the quality of the high and low decision.
Inline assembly is passed in C + +
actually found the 500W data sorting results are as follows:
Algorithm name inline assembler algorithm time C + + algorithm time
Bubble sort 5W data slow to die 5W data slowly dying.
Quick sort 600ms about 500ms around
------------------Why there is a fast sorting algorithm, the compiled results are not as high as C/s + + efficiency, because I write inline assembly without compiler automatic generation of high efficien
body | ---------> @ 2|------> | -------------------- +|| Decryptor || ---------> @ 3+ -------------------- +@ 1 is a call constructed by computation, because the call location must be determined by @ 2.@ 2 is an encrypted virus.@ 3 is an encryptor used to decrypt @ 2, which is transformed by code obfuscation.In this way, every time other files are infected, the re-generated code will no longer have a fixed feature, which will invalidate the feature scanning mechanism.
2.1 random number design:T
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