CHG: syntax sugar, minor spelling mistake
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@@ -77,7 +77,7 @@ static int debug_print = 0;
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* @brief The key diversification algorithm uses 6-bit bytes.
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* This implementation uses 64 bit uint to pack seven of them into one
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* variable. When they are there, they are placed as follows:
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* XXXX XXXX N0 .... N7, occupying the lsat 48 bits.
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* XXXX XXXX N0 .... N7, occupying the last 48 bits.
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*
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* This function picks out one from such a collection
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* @param all
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@@ -141,38 +141,28 @@ uint64_t swapZvalues(uint64_t c)
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*/
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uint64_t ck(int i, int j, uint64_t z)
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{
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if(i == 1 && j == -1)
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{
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if (i == 1 && j == -1) {
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// ck(1, −1, z [0] . . . z [3] ) = z [0] . . . z [3]
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return z;
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}else if( j == -1)
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{
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} else if( j == -1) {
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// ck(i, −1, z [0] . . . z [3] ) = ck(i − 1, i − 2, z [0] . . . z [3] )
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return ck(i-1,i-2, z);
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}
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if(getSixBitByte(z,i) == getSixBitByte(z,j))
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if (getSixBitByte(z,i) == getSixBitByte(z,j))
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{
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//ck(i, j − 1, z [0] . . . z [i] ← j . . . z [3] )
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uint64_t newz = 0;
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int c;
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for(c = 0; c < 4 ;c++)
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{
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for(c = 0; c < 4; c++) {
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uint8_t val = getSixBitByte(z,c);
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if(c == i)
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{
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if (c == i)
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pushbackSixBitByte(&newz, j, c);
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}else
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{
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else
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pushbackSixBitByte(&newz, val, c);
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}
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}
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return ck(i,j-1,newz);
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}else
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{
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} else {
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return ck(i,j-1,z);
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}
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}
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@@ -364,9 +354,9 @@ void hash0(uint64_t c, uint8_t k[8])
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k[i] |= zTilde_i & 0x7E;
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k[i] |= (~p_i) & 1;
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}
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if((k[i] & 1 )== 0)
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if ((k[i] & 1 )== 0)
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{
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zerocounter ++;
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zerocounter++;
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}
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}
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}
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@@ -694,10 +684,8 @@ int readKeyFile(uint8_t key[8]) {
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if ( bytes_read == 1)
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retval = 0;
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if (f) {
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if (f)
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fclose(f);
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f = NULL;
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}
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return retval;
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}
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