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Yarbay Konu Sahibi 13 yıl 0 64 bitde oyunu bırakın açmayı yüklenmiyor bile. Bunu bir çözümü var mıdır ? Sonraki Mesaj Yüzbaşı 13 yıl 0 DOSBox ile açmayı denedin mi? Sonraki Mesaj Buna gelen 1 yanıtı gör. Yüzbaşı 13 yıl 0 64bit desteklemiyor olabilir Sonraki Mesaj Buna gelen 1 yanıtı gör. Binbaşı 13 yıl 0 Uyumluluk sorunu var , uyumluluk modunda Xp falan seç olmassa yapıcak birşey yok. Sonraki Mesaj Buna gelen 1 yanıtı gör. Yapay Zeka’dan İlgili Konular Need for Speed Underground 2 KristalTürk, 4 ay önce ÇİFT KİŞİ OYNANABİLEN NEED FOR SPEED SÜRÜMÜ??? Astentra, 12 yıl önce Need for Speed: Carbon (2006) [ANA KONU] yorkashton, 3 yıl önce Need for Speed Underground, RTX Remix ile yepyeni bir oyun oldu sniper23, 8 ay önce Need For Speed Underground 1'in save dosyası nerde? ahbabamsağolsaydı, 12 yıl önce NEED FOR SPEED 2 SE 3DFX destegi ile oynama.. cesim akyıldız, 7 yıl önce Yarbay 13 yıl 0 ben de yaşadım aynı sorunu dosboxla oynanıyormuş ama çok zahmetli geldi nostalji de yalan oldu Sonraki Mesaj Buna gelen 1 yanıtı gör. Yarbay Konu Sahibi 13 yıl 0 öneriler için teşekkürler deneyeceğim. Sonraki Mesaj Buna gelen 1 yanıtı gör. Yarbay 13 yıl 0 Origin installer ile kur crack atarken önceki Need For Speed The Run.exe yi silme üzerine at sorun düzelicektir benimkide 64 bit Sonraki Mesaj Buna gelen 2 yanıtı gör. Binbaşı 13 yıl 0 Sonraki Mesaj Yarbay 13 yıl 0 quote:Orijinalden alıntı: Mert-TR Origin installer ile kur crack atarken önceki Need For Speed The Run.exe yi silme üzerine at sorun düzelicektir benimkide 64 bit Konuyu okusaymışsın iyiymiş Sonraki Mesaj Buna gelen 1 yanıtı gör. Yarbay 13 yıl 0 quote:Orijinalden alıntı: adilazimdegilx quote:Orijinalden alıntı: Mert-TR Origin installer ile kur crack atarken önceki Need For Speed The Run.exe yi silme üzerine at sorun düzelicektir benimkide 64 bit Konuyu okusaymışsın iyiymiş Benim kafa Shift e gitmiş yeni fark ettim --> --> Sonraki Mesaj . Avocode 3.0.0 (32-bit) Avocode 4.15.6 (64-bit) Top Downloads. Malwarebytes Malwarebytes ; Mailbird Mailbird .0; Opera Opera 98.0 Build 4759.15 (64-bit) Photoshop Adobe

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Sponsored ContentThis sponsored post features a product relevant to our readers while meeting our editorial guidelines for being objective and educational.In today’s tutorial I’m going to introduce you to Avocode, developed by the “eleven brave men and one brave woman” of Source and recently released in the form of version 1.0. Avocode is an application which allows you to work with .psd and .sketch files, moving from design to code, without even opening Photoshop or Sketch.We covered Avocode a few months ago, but back then it was still in beta and has since undergone some serious feature overhauls. This tutorial will show you how to work with the Avocode desktop application and introduce the new features. Part 1. Getting StartedIn this part of the tutorial I’ll demonstrate how to get setup with Avocode, create projects, upload Photoshop and Sketch files, then work with those files in the Avocode desktop application. Part 2. Exporting CodeIn this video I’ll show you how to generate CSS (along with Sass and Compass) directly from your Photoshop and Sketch files. We’ll also look at some shortcuts and a ton of other helpful features. Let’s get started! Useful Resourcesavocode.com@avocode on TwitterAvocode Stories on Medium

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And 64-bit result (L) is obtained. Second, the multiplication is performed with lower 32-bit of 64-bit operands ( A [ 7 ∼ 4 ] , B [ 7 ∼ 4 ] ) and 64-bit result (H) is obtained. Third, XOR operation is performed between the upper part and the lower part of the A and B, respectively. Values are multiplied and output the final result (M). The L, H, and M are XORed together and stored in M, again. Lastly, shifted H ( H ≪ 64 ), shifted M ( M ≪ 32 ) and L are XORed and the final 128-bit result (C) is obtained.Karatsuba algorithm needs a buffer to store intermediate results of each operation. In the Block Comb method proposed by Seo and Kim, the size of intermediate results is over than given size of general purpose registers. For this reason, the part of intermediate result should be stored in STACK. However, accessing to STACK memory (i.e., 2 clock cycles) requires high-overheads than register accesses (i.e., 1 clock cycle). The proposed method uses eight more registers to maintain the intermediate result. For the register optimized version, the intermediate result is stored in STACK memory.In Algorithm 2, the proposed implementation does not reserve the intermediate result of the Karatsuba calculation, such as L, H, and M, in the STACK. Instead, only values needed for the following operation are stored using the K 0 to K 7 registers. When the 128-bit result (C) and the 64-bit intermediate result (M) are divided into 8-bit units, it can be expressed as ( C [ 0 ] , C [ 1 ] , . . . , C [ 15 ] ) and ( M [ 0 ] , M [ 1 ] , . . . , M [ 8 ] ), respectively. Among them, C [ 0 ∼ 3 ] and C [ 12 ∼ 15 ] store the upper 32-bit multiplication result and the lower 32-bit multiplication result.The result from C [ 4 ] to C [ 11 ] is XORed with the M. The final result of C

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[ 4 ] is calculated by C [ 4 ] ⨁ M [ 0 ] ⨁ C [ 0 ] ⨁ C [ 8 ] . Values affected by XOR, such as C [ 0 ] ∼ C [ 3 ] and C [ 12 ] ∼ C [ 15 ] , are stored after each 32-bit multiplication operation. In addition, values used to make C [ 4 ] ∼ C [ 11 ] can be XORed and stored in advance. Afterward, 64-bit multiplication is performed without loading and storing procedures from the STACK.In the case of the GHASH function, the associated data and cipher text are XORed and multiplied by the hash key value ( H G H A S H ). In this process, the B value, which is a multiplier of all 128-bit multiplication operations, is fixed to the H G H A S H value. This repeated value (i.e., H G H A S H [ 7 ∼ 0 ] ⨁ H G H A S H [ 3 ∼ 0 ] ) can be pre-computed and reserved to reduce the number of XOR operations and memory accesses. 4.3. Fast Reduction for GHASHIn order to optimize the modular operation, the fast reduction modulo algorithm by Gueron and Kounavis [16] was applied to the proposed implementation. The algorithm optimizes the reduction operation with an irreducible polynomial ( x 128 + x 7 + x 2 + x + 1 ) in a 64-bit environment. The algorithm is optimized for 8-bit AVR platforms. In Algorithm 3, input values ( K [ 0 ] ∼ K [ 31 ] ) were reflected in advance. This is the reason why all of operations are reversed to the existing original algorithm. The algorithm requires a number of bit-wise shift operation. The 8-bit wise shift operation is optimized away. Algorithm 3 Proposed method for fast modular reduction. Input:: 256-bit data K [ 0 ] , . . . , K [ 31 ] , where K [ 0 ∼ 31 ] are 8-bit long each. Output:: K [ 16 ] , .

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The example above, these will be: d7ce56a3 566656f3.Let’s reverse the order of bytes in each 4-byte sequences and form two 32-bit integer numbers: a356ced7 f3566656.Together these two sequences form a 64-bit integer number: a356ced7f3566656.This number in the binary format will look as following:0xA356CED7F3566656 = 1010001101010110110011101101011111110011010101100110011001010110Let’s divide the 64-bit value into groups of 5-bit each starting from least significant bits (from right to left):there will be 12 groups 5-bit each, and the last group will have remaining 4-bit.Each 5-bit group can be represented by a 5-bit integer number in the 0-31 range:1010 00110 10101 10110 01110 11010 11111 11001 10101 01100 11001 10010 10110 10 6 21 22 14 26 31 25 21 12 25 18 22Finally, we are going to encode each 5-bit number into a single character using this lookup table:'0','1','2','3','4','5','6','7','8','9','a','b','c','d','f','g','h','j','k','m','n','p','q','r','s','t','u','v','w','x','y','z'Here is a final result, 13-character string, used in Oracle as SQL_ID: 10 6 21 22 14 26 31 25 21 12 25 18 22 a 6 p q f u z t p c t k qSQL_ID = 'a6pqfuztpctkq'Let’s check that with the following SQL and results in Figure 3.select ADDRESS, SQL_ID, HASH_VALUE, to_char(HASH_VALUE,'FMXXXXXXXX') HASH_VALUE_HEXfrom v$sqlwhere sql_text='select 0 from dual';Let’s make few practical conclusions.The possibility of SQL_ID collisions is significantly lower (about 4 billion times lower) than for HASH_VALUE.That means much stronger one-to-one relationship between SQL text and SQL_ID.Therefore, the same SQL_ID means the same SQL text.It no longer matters if they were on different databases, different versions, or different platforms.In this case, same text means that both. Avocode 3.0.0 (32-bit) Avocode 4.15.6 (64-bit) Top Downloads. Malwarebytes Malwarebytes ; Mailbird Mailbird .0; Opera Opera 98.0 Build 4759.15 (64-bit) Photoshop Adobe

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Go, but if the data isn’t aligned, it might take multiple operations to access it.Alignment issuesThings get tricky on 32-bit architectures, the compiler doesn’t guarantee that 64-bit values will be aligned on an 8-byte boundary. Instead, they might only be aligned on a 4-byte boundary.This becomes a problem when we use the atomic package to perform operations on the state variable. The atomic package specifically notes:“On ARM, 386, and 32-bit MIPS, it is the caller’s responsibility to arrange for 64-bit alignment of 64-bit words accessed atomically via the primitive atomic functions.”Therefore, if we don’t align the state uint64 variable to an 8-byte boundary on these 32-bit architectures, it could cause the program to crash.So, what’s the fix? Let’s take a look at how this has been handled across different versions.Go 1.5: state1 [12]byteI’d recommend taking a moment to guess the underlying logic of this solution as you read the code below, then we’ll walk through it together.type WaitGroup struct { state1 [12]byte sema uint32}func (wg *WaitGroup) state() *uint64 { if uintptr(unsafe.Pointer(&wg.state1))%8 == 0 { return (*uint64)(unsafe.Pointer(&wg.state1)) } else { return (*uint64)(unsafe.Pointer(&wg.state1[4])) }}Instead of directly using a uint64 for state, WaitGroup sets aside 12 bytes in an array (state1 [12]byte). This might seem like more than you’d need, but there’s a reason behind it.WaitGroup in Go 1.5The purpose of using 12 bytes is to ensure there’s enough room to find an 8-byte segment that’s properly aligned.The alignment of the struct depends on its field’s type, so it is either 8-byte or