Problems in second blocks message digest in self-learning SHA-1 algorithm - hash

Question

Output
I am new in learning C programming. Now, I am trying to do SHA-1 for university project. I think this coding by myself. I am trying to do the message digest from the file above 55characters, which means 2 blocks is needed. The first block message digest is correct, but the second block is wrong. I have checked it very many times, but I still not able to find the mistake. Can anyone with experiences able to help me find it out? Thank you.
patients information.txt
"abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int length,number_of_block,str_length;
unsigned int i = 0,j = 0,l = 0,e = 0,n = 0, t=0, k=0,f=0;
float x=0;
char c;
int H[5]={0x67452301,0xEFCDAB89,0x98BADCFE,0x10325476,0xC3D2E1F0};
unsigned int temp = 0;
FILE *file;
file = fopen("patients information.txt", "r");//Choose the file that want to access
if (file == NULL)//detect the file is empty or not
{
printf("The file is empty");
}
fseek(file, 0, SEEK_END);// move the file pointer to the end of the file
length = ftell(file);//calculate the length of sting in file
fseek(file, 0, SEEK_SET);// move file pointer back to start of file so we can read each character
printf("The length of string is %d\n",length);//check the number of character in string
char *string = malloc(sizeof(char) * (length+1));
while ( (c = fgetc(file)) != EOF)//pass the every character in the file to the string array
{
string[i] = c;
i++;
}
string[i] = '\0';//terminate the string storing
unsigned char long_msg[length+1];
for (i=0;i<length;i++)//pass the pointer array to the unsigned character array
{
long_msg[i]=string[i];
}
printf("The character store in th array is ");
long_msg[length]=128;
for (i=0;i<=length;i++)//check the message in msg array
{
printf("%X ",long_msg[i]);
}
if (length<=55)
{
number_of_block = 1;
}
else if (length>55 && length<120)
{
number_of_block = 2;
}
else
{
x = ((length - 55)/64);//calculate the number of block needed
number_of_block = x+2;
}
printf("\nNumber of block needed is %d\n",number_of_block);
unsigned char blocks[number_of_block][64];
for (i=0;i<number_of_block;i++)//Split the long string into n number of blocks
{
for(j=0;j<64;j++)
{
blocks[i][j]=long_msg[l];
if(l>length)//padding 0
{
blocks[i][j]=0;
}
l++;
}
}
for (i=0;i<number_of_block;i++)//check the blocks content
{
for(j=0;j<64;j++)
{
printf("%X ",blocks[i][j]);
}
}
printf("\nCheck length padding:\n");
str_length = 8*length;//sting length in bits
if (length<32)//if length of string is 1 bytes in hexadecimal
{
blocks[number_of_block-1][63]=str_length;
}
else
{
blocks[number_of_block-1][62]=(str_length>>8);//second last block
blocks[number_of_block-1][63]=((str_length<<8)>>8);//last block
}
for (i=0;i<number_of_block;i++)//check length padding
{
for(j=0;j<64;j++)
{
printf("%02X ",blocks[i][j]);
}
}
unsigned int w[number_of_block][16][4];
unsigned int W[number_of_block][80];
unsigned int A[number_of_block],B[number_of_block],C[number_of_block],D[number_of_block],E[number_of_block];
for (e=0;e<number_of_block;e++)
{
/*The problem is here*/
n=0;
for (i=0;i<16;i++)//split the padding message into w0 to w15 ,exp. w0 = (w[0][1]),....,(w[0][3])
{
for(j=0;j<4;j++)
{
w[e][i][j] = blocks[e][n];
n++;
}
}
for (i=0;i<16;i++)//combine the hex --> 16 block of hexadecimal(W0 to W15)
{
W[e][i] = ((w[e][i][0])<<24 | (w[e][i][1])<<16 | (w[e][i][2])<<8 | (w[e][i][3]));
}
/*Compute message digest*/
A[e] = 0x67452301;
B[e] = 0xEFCDAB89;
C[e] = 0x98BADCFE;
D[e] = 0x10325476;
E[e] = 0xC3D2E1F0;
for (t=0;t<=79;t++)
{
//for t = 16 -> 79
if (t>=16 && t<=79)//prepare W16 to W79
{
W[e][t]= ( (W[e][t-3]) ^ (W[e][t-8]) ^ (W[e][t-14]) ^ (W[e][t-16]) );
W[e][t]= ( ((W[e][t])<<1) | ((W[e][t]) >> (32-1)));//perform circular left shift
}
if (t>=0 && t<=19)
{
f = (B[e]&C[e]) | ((~B[e])&D[e]);
k = 0x5A827999;
}
else if (t>=20 && t<=39)
{
f = (B[e]^C[e]^D[e]);
k = 0x6ED9EBA1;
}
else if (t>=40 && t<=59)
{
f = (B[e]&C[e]) | (B[e]&D[e]) | (C[e]&D[e]);
k = 0x8F1BBCDC;
}
else if(t>=60 && t<=79)
{
f = (B[e]^C[e]^D[e]);
k = 0xCA62C1D6;
}
temp = ((A[e]<<5) | (A[e] >> (32-5))) + f + E[e] + W[e][t] + k;
E[e] = D[e];
D[e] = C[e];
C[e] = ( (B[e]<<30) | (B[e]>> (32-30)));
B[e] = A[e];
A[e] = temp;
}
printf("\n\n");
printf("%08X %08X %08X %08X %08X",A[e],B[e],C[e],D[e],E[e]);//check the value before adding up
H[0] = ( H[0] + A[e]);
H[1] = ( H[1] + B[e]);
H[2] = ( H[2] + C[e]);
H[3] = ( H[3] + D[e]);
H[4] = ( H[4] + E[e]);
}
printf("\n\n");
printf("Message digest:");
for (i=0;i<5;i++)
{
printf("%X ",H[i]);
}
printf("\n\n");
return 0;
}
The wrong output of second block : CE3A1FD0 01464A63 F6766B50 AF97AC62 8D5DBBDD
The output of second block should be: 906FD62C 58C0AAC0 B6A55520 74E9B89D 9AF00B7F