pbkdf2 c语言实现

#include <math.h>
#include <stdio.h>
#include <stdlib.h>


unsigned long H[5];
unsigned long T[512]={0};


unsigned long f(unsigned long B,unsigned long C,unsigned long D, int t)
{
	if (t < 20)
		{
			return ((B & C)^((~B) & D));
		}
	if ((t > 19) & (t < 40))
		{
			return (B ^ C ^ D);
		}
	if ((t > 39) & (t < 60))
		{
			return ((B & C)^(B & D)^(C & D));
		} 
	if (t > 59)
		{
			return (B ^ C ^ D);
		}
	return B;
}




unsigned long Rol(unsigned long x, int y)
{
	if (y % 32 == 0) {return x;} 
	else {return ((x << y)^(x >> -y));}
}


unsigned long Ror(unsigned long x, int y)
{
	if (y % 32 == 0) {return x;}
	else {return ((x >> y)^(x << -y));}
}


void append(char* s, int slen, char c)
{
        //int len = strlen(s);
        s[slen] = c;
        s[slen+1] = '\0';
}


void SHA1(char s[],int slen)
{
	unsigned long K[80];
	unsigned long A,B,C,D,E,TEMP;
	int r,k,ln,t,l,i,j;
	H[0]=0x67452301;
	H[1]=0xefcdab89;
	H[2]=0x98badcfe;
	H[3]=0x10325476;
	H[4]=0xc3d2e1f0;


	ln=slen;
	r = (int)((ln+1)/64);


	if (((ln+1) % 64) > 56)
		{
		r=r+1;
		}


	// initialize Constants


	for(t=0; t<80; t++)
		{
			if (t<20)
				{
					K[t] = 0x5a827999;
				}


			if ((t>19)&(t<40))
				{
					K[t] = 0x6ED9EBA1;
				}
			if ((t>39)&(t<60))
				{
					K[t] = 0x8F1BBCDC;
				}
			if (t>59)
				{
					K[t] = 0xca62c1d6;
				}
		}


	for(l=0; l <= r; l++)
	{
		unsigned long W[80]={0};
		//Initialize Text
		for (i=0; i<16; i++)
			{
			for(j=0; j<4; j++)
				{
					if (4*i+j <= ln)
					{
						k = s[64*l+4*i+j];
					}
					else
					{
						k =0;
					}
				
					if (k<0)
					{
						k = k +256;
					}


					if (4*i+j == ln)
						{
							k = 0x80;
						}


					W[i]= W[i] + k*(unsigned long)pow(256,(double)3-j);
				}
			}
		if ((W[14]==0)&(W[15]==0))
		{
			W[15]=8*slen;
		}


	// Hash Cycle


		for (t = 16; t <80; t++)
			{
				W[t] = Rol(W[t-3]^W[t-8]^W[t-14]^W[t-16],1);
			}


		A = H[0];
		B = H[1];
		C = H[2];
		D = H[3];
		E = H[4];


		for(t = 0; t < 80; t++)
		{
			TEMP = Rol(A,5) + f(B,C,D,t) + E + W[t] + K[t];
			E = D;
			D = C;
			C = Rol(B,30);
			B = A;  
			A = TEMP;
		}


		H[0] = H[0] + A;
		H[1] = H[1] + B;
		H[2] = H[2] + C;
		H[3] = H[3] + D;
		H[4] = H[4] + E;


		ln = ln - 64;
	}


} 


void HMAC(char text[], int tlen, char key[], int klen)
{
	char c;
	char s[256] = "";
	int slen = 0;
	unsigned long Key[16] = {0};
	unsigned long X[16] = {0};
	unsigned long Y[16] = {0};
	unsigned long ipad = 0x36363636;
	unsigned long opad = 0x5c5c5c5c;
	int k,i,j;




	//Process string key into sub-key
	//Hash key in case it is less than 64 bytes
	if (klen > 64)
	{
		SHA1(key, klen);
		Key[0] = H[0];
		Key[1] = H[1];
		Key[2] = H[2];
		Key[3] = H[3];
		Key[4] = H[4];
	}
	else
	{
		for(i=0; i<16; i++)
		{
			for(j=0; j<4; j++)
			{
				if (4*i+j <= klen)
				{
					k = key[4*i+j];
				}
				else
				{
					k = 0;
				}
				if (k<0)
				{
					k = k + 256;
				}
				Key[i]= Key[i] + k*(unsigned long)pow(256,(double)3-j);
			}
		}
	}
	
	for(i=0; i<16; i++)
	{
		X[i] = Key[i]^ipad;
		Y[i] = Key[i]^opad;
	}


	//Turn X-Array into a String
	for(i=0; i<16; i++)
	{
		for(j=0; j<4; j++)
		{
			c = ((X[i] >> 8*(3-j)) % 256);
			//s = s + c;
			append(s,slen,c);
			slen++;
		}
	}


	//Append text to string
	//s = s + text
	for (i=0;i<tlen;++i)
	{
		s[slen+i] = text[i];
	}
	slen = slen + tlen;


	//Hash X-Array
	SHA1(s,slen);


	for (i=0;i<slen;++i)
	{
		s[i] = 0;
	}
	slen = 0;


	//Turn Y-Array into a String
	for(i=0; i<16; i++)
	{
		for(j=0; j<4; j++)
		{
			c = ((Y[i] >> 8*(3-j)) % 256);
			//s = s + c;
			append(s,slen,c);
			slen++;
		}
	}


	//Append Hashed X-Array to Y-Array in string
	for(i=0; i<5; i++)
	{
		for(j=0; j<4; j++)
		{
			c = ((H[i] >> 8*(3-j)) % 256);
			//s = s + c;
			append(s,slen,c);
			slen++;
		}
	}


	//Hash final concatenated string
	SHA1(s,slen);
}
void PBKDF2(char P[],int Plen, char S[], int Slen, int c,int dkLen)
{
	unsigned long L[5]={0,0,0,0,0};
	int hlen,l,r,i,j,x,y;
	char t;
	
	hlen = 20;							// from sha1
	//requires new version of Ceiling Function: l = ceil(dkLen/hlen)
	l = (dkLen/hlen)+1;					// hLen-octets 32/20 = 2 always
	r = dkLen -(l-1)*hlen;				// number of octets in the last block (32-20) = 12
		
	for(i=1; i<=l; i++)					// passes
	{
		
		// t = 0;
		// U = S + t + t + t ;				// append characters
		// t = i;
		// U = U + t;
		/////////
		char U[50];  // larger than it needs to be (SSID<32 + 4)
		int Ulen = 0;
		char appended[] = {(char)0, (char)0, (char)0, (char)i};
 
		for (j=0;j<Slen;++j)
		{
			U[j]=S[j];
		}
		for (j=0;j<5;++j)
		{
			U[14+j] = appended[j];
		}
		Ulen = Slen + 4;
		/////////
		//
		//
		//
		//


		L[0] = L[1] = L[2] = L[3] = L[4] = 0;
		for(j=1; j<=c; j++)			// number of iterations (4096)
		{


			HMAC(U,Ulen,P,Plen);
			L[0] = L[0]^H[0];		// ^ is the XOR function
			L[1] = L[1]^H[1];
			L[2] = L[2]^H[2];
			L[3] = L[3]^H[3];
			L[4] = L[4]^H[4];
			
			//
			// U = "";
			/////////
			for (x=0;x<Ulen;++x)
			{
				U[x] = 0;
			}
			Ulen=0;
			/////////
			//
			//


			for(x=0; x<5; x++)
			{
				for(y=0; y<4; y++)
				{
					t = ((H[x] >> 8*(3-y)) % 256);
					
					//
					// U = U + t;
					/////////
					append(U,Ulen,t);
					Ulen++;
					/////////
					//
					//
				}
			}
		}
		
		T[5*(i-1)] = L[0];  // moves the results down 5 each iteration
		T[5*(i-1)+1] = L[1];
		T[5*(i-1)+2] = L[2];
		T[5*(i-1)+3] = L[3];
		T[5*(i-1)+4] = L[4];
	}
}


int main()
{ 
	int i,j;
	printf("\n\n****---***---***---***---***---***---****\n\n");
	SHA1("abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq",56);
	printf("SHA1 tests (Test Cases from FIPS PUBS 180-2)\n");
	printf("\nText     : abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq\n");
	printf("Expected : 84983e44 1c3bd26e baae4aa1 f95129e5 e54670f1\n");
	printf("Returned :");
	for(i=0;i<=4;++i)
	{
		printf(" %x",H[i]);
	}
	SHA1("abc",3);
	printf("\n\nText     : abc\n");
	printf("Expected : a9993e36 4706816a ba3e2571 7850c26c 9cd0d89d\n");
	printf("Returned :");
	for(i=0;i<=4;++i)
	{
		printf(" %x",H[i]);
	}
	printf("\n\n****---***---***---***---***---***---****\n\n");
	HMAC("what do ya want for nothing?",28,"Jefe",4);
	printf("HMAC test (Test Case from RFC2202)\n\n");
	printf("Key      : Jefe\n");
	printf("Text     : what do ya want for nothing?\n");
	printf("Expected : effcdf6a e5eb2fa2 d27416d5 f184df9c 259a7c79\n");
	printf("Returned :");
	for(i=0;i<=4;++i)
	{
		printf(" %x",H[i]);
	}
	printf("\n\n****---***---***---***---***---***---****\n\n");
	// use http://www.xs4all.nl/~rjoris/wpapsk.html
	// and http://anandam.name/pbkdf2/
	// for PBKDF2 Verification
	PBKDF2("Password22",10,"itsMEjessica22",14,4096,32);
	printf("PBKDF2 test (Test Cases from internet, see comments)\n");
	printf("SSID     : itsMEjessica22\n");
	printf("Password : Password22\n");
	printf("Expected : 2944bf07 17219278 f0e990ac 34e4e151 1c4233ae b38b63cc f4d39035 3426233e\n");
	printf("Returned :");
	for(j=0; j<(32/4); j++)
	{
		printf(" %x",T[j]);
	}
}