Two questions with base64 encoding - visual-c++

I confused how to convert const char * to base64 with 2 Questions:
Question #1 how do I defined the length of output string that would perfectly match the length of output base64?I have found a code which from apple opensource,the code in below http://www.opensource.apple.com/source/QuickTimeStreamingServer/QuickTimeStreamingServer-452/CommonUtilitiesLib/base64.c
or I could directly use "atlenc.h" in VC++.if the length of coded_dst which I have defined is smaller than the actually,the program may crashed
int Base64encode(char *coded_dst, const char *plain_src, int len_plain_src)
{
const char basis_64[] ="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
int i;
char *p;
p = coded_dst;
for (i = 0; i < len_plain_src - 2; i += 3) {
*p++ = basis_64[(plain_src[i] >> 2) & 0x3F];
*p++ = basis_64[((plain_src[i] & 0x3) << 4) |
((int) (plain_src[i + 1] & 0xF0) >> 4)];
*p++ = basis_64[((plain_src[i + 1] & 0xF) << 2) |
((int) (plain_src[i + 2] & 0xC0) >> 6)];
*p++ = basis_64[plain_src[i + 2] & 0x3F];
}
if (i < len_plain_src) {
*p++ = basis_64[(plain_src[i] >> 2) & 0x3F];
if (i == (len_plain_src - 1)) {
*p++ = basis_64[((plain_src[i] & 0x3) << 4)];
*p++ = '=';
}
else {
*p++ = basis_64[((plain_src[i] & 0x3) << 4) |
((int) (plain_src[i + 1] & 0xF0) >> 4)];
*p++ = basis_64[((plain_src[i + 1] & 0xF) << 2)];
}
*p++ = '=';
}
*p++ = '\0';
return p - coded_dst;
}
Question #2 as we all well know that the type of byte in C++ is unsigned char,how do I convert the char * to unsigned char *?
thanks
regards
Ken

The design of your function, based on the signature, tells me it's up to the caller to provide a sufficient buffer for output. This would be unsafe in your example because the caller isn't informing the function how large that buffer is. Your function has no chance to limit output to coded_dst to the buffer provided, so you should add, at the least, a parameter for that.
As such, you would need to check as you loop to be sure p, a pointer into coded_dst, stays within that limit, returning an error to the caller if there's insufficient room.
That said, notice how many increments of p occur for every 3 source items processed. The ratio is 3/4...for every 3 that go into that loop, 4 come out. So, to start the calculation of the required length, begin with
( len_plain_src / 3 ) * 4;
Now, consider r = len_plain_src % 3; If r is zero, your algorithm adds 2 more bytes. If r has a remainder, your algorithm adds 3 more bytes.
After that, you append a zero terminator.
Look carefully, I've not clearly analyzed this, but you may have an error in the closing '=' appended at the tail for the case where (i<len_plain_src) - you may have added two of them instead of just one.
Now, to handle the unsigned char, you could change the declaration and initial assignment of p with,
unsigned char * p = (unsigned char *) coded_dst;
At which point it would be more convenient for you if you declare basis_64 to be unsigned char

Related

Best data type and rounding function for weight and currency variables

I need to multiply two values ​​- weight and currency (Visual c++, mfc). E.g.:
a=11.121;
b=12.11;
c=a*b;
Next I have to round "с" to 2 digits after point (currency value, e.g. 134.68). What the best data types and rounding function for this variables? The rounding procedure must be mathematically correct.
P.S. The problem was solved by very ugly but working part of code:
CString GetPriceSum(CString weight,CString price)
{
price.Replace(".", "");
price = price + "0";
if (weight.Find(".") == -1) { weight = weight + ".000"; }
weight.Replace(".", "");
unsigned long long int iprice = atoi(price);
unsigned long long int iweight = atoi(weight);
unsigned long long int isum = iprice * iweight;
CString sum = ""; sum.Format("%llu", isum);
CString r1 = sum.Right(1);
if (atoi(r1) >= 5) { isum += 10; }
CString r2 = sum.Mid(sum.GetLength() - 2, 1);
if (atoi(r2) >= 5) { isum += 100; sum.Format("%llu", isum);}
r2 = sum.Mid(sum.GetLength() - 3, 1);
if (atoi(r2) >= 5) { isum += 1000; sum.Format("%llu", isum);}
r2 = sum.Mid(sum.GetLength() - 4, 1);
if (atoi(r2) >= 5) { isum += 10000; sum.Format("%llu", isum);}
CString finsum = ""; finsum.Format("%llu", isum);
finsum.Insert(finsum.GetLength() - 6, ".");
finsum.Delete(finsum.GetLength() - 4, 4);
if (finsum.Left(1) == ".") { finsum = "0" + finsum; }
return finsum;
}
How about this: let's start from
API I use, counts values using some other language. And they round they values mathematically correct.
In your other question, you got those value as strings. You can construct an integer from those digits (remove decimal point). Assuming that the product fits in a 64-bit int, you can multiply them exactly. Now you can manually round to a desired precision and drop unneeded digits.
Code example (you may want to add error checking):
#define _CRT_SECURE_NO_WARNINGS
#include <string>
#include <iostream>
#include <sstream>
int main()
{
std::string a = "40.50";
std::string b = "0.490";
long long l1, dec1, l2, dec2;
sscanf(a.data(), "%lld.%lld", &l1, &dec1);
l1 = l1 * 100 + dec1;
sscanf(b.data(), "%lld.%lld", &l2, &dec2);
l2 = l2 * 1000 + dec2;
long long r = l1 * l2;
r /= 100;
int rem = r % 10;
r /= 10;
if (rem >= 5)
r++;
std::stringstream ss;
ss << r / 100 << "." << std::setw(2) << std::setfill('0') << r % 100;
std::cout << ss.str();
}
You can also use stringstream instead of sscanf to parse the strings.

Generate Checksum for String

I would like to Generate Checksum for Strings/Data
1. The same data should produce the same Checksum
2. Two different data strings can't product same checksum. Random collision of 0.1% can be negligible
3. No encryption/decryption of data
4. Checksum length need not be too huge and contains letters and characters.
5. Must be too fast and efficient. Imagine generating checksum(s) for 100 Mb of text data should be in less than 5mins. Generating 1000 checksums for less than 1 KB of each segment data should be in less than 10 seconds.
Any algorithm or implementation reference and suggestions are most appreciated.
You can write a custom hash function: (c++)
long long int hash(String s){
long long k = 7;
for(int i = 0; i < s.length(); i++){
k *= 23;
k += s[i];
k *= 13;
k %= 1000000009;
}
return k;
}
This should give you a well (collision free for most samples) hash value.
A very common, fast checksum is the CRC-32, a 32-bit polynomial cyclic redundancy check. Here are three implementations in C, which vary in speed vs. complexity, of the CRC-32: (This is from http://www.hackersdelight.org/hdcodetxt/crc.c.txt)
#include <stdio.h>
#include <stdlib.h>
// ---------------------------- reverse --------------------------------
// Reverses (reflects) bits in a 32-bit word.
unsigned reverse(unsigned x) {
x = ((x & 0x55555555) << 1) | ((x >> 1) & 0x55555555);
x = ((x & 0x33333333) << 2) | ((x >> 2) & 0x33333333);
x = ((x & 0x0F0F0F0F) << 4) | ((x >> 4) & 0x0F0F0F0F);
x = (x << 24) | ((x & 0xFF00) << 8) |
((x >> 8) & 0xFF00) | (x >> 24);
return x;
}
// ----------------------------- crc32a --------------------------------
/* This is the basic CRC algorithm with no optimizations. It follows the
logic circuit as closely as possible. */
unsigned int crc32a(unsigned char *message) {
int i, j;
unsigned int byte, crc;
i = 0;
crc = 0xFFFFFFFF;
while (message[i] != 0) {
byte = message[i]; // Get next byte.
byte = reverse(byte); // 32-bit reversal.
for (j = 0; j <= 7; j++) { // Do eight times.
if ((int)(crc ^ byte) < 0)
crc = (crc << 1) ^ 0x04C11DB7;
else crc = crc << 1;
byte = byte << 1; // Ready next msg bit.
}
i = i + 1;
}
return reverse(~crc);
}
// ----------------------------- crc32b --------------------------------
/* This is the basic CRC-32 calculation with some optimization but no
table lookup. The the byte reversal is avoided by shifting the crc reg
right instead of left and by using a reversed 32-bit word to represent
the polynomial.
When compiled to Cyclops with GCC, this function executes in 8 + 72n
instructions, where n is the number of bytes in the input message. It
should be doable in 4 + 61n instructions.
If the inner loop is strung out (approx. 5*8 = 40 instructions),
it would take about 6 + 46n instructions. */
unsigned int crc32b(unsigned char *message) {
int i, j;
unsigned int byte, crc, mask;
i = 0;
crc = 0xFFFFFFFF;
while (message[i] != 0) {
byte = message[i]; // Get next byte.
crc = crc ^ byte;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >> 1) ^ (0xEDB88320 & mask);
}
i = i + 1;
}
return ~crc;
}
// ----------------------------- crc32c --------------------------------
/* This is derived from crc32b but does table lookup. First the table
itself is calculated, if it has not yet been set up.
Not counting the table setup (which would probably be a separate
function), when compiled to Cyclops with GCC, this function executes in
7 + 13n instructions, where n is the number of bytes in the input
message. It should be doable in 4 + 9n instructions. In any case, two
of the 13 or 9 instrucions are load byte.
This is Figure 14-7 in the text. */
unsigned int crc32c(unsigned char *message) {
int i, j;
unsigned int byte, crc, mask;
static unsigned int table[256];
/* Set up the table, if necessary. */
if (table[1] == 0) {
for (byte = 0; byte <= 255; byte++) {
crc = byte;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >> 1) ^ (0xEDB88320 & mask);
}
table[byte] = crc;
}
}
/* Through with table setup, now calculate the CRC. */
i = 0;
crc = 0xFFFFFFFF;
while ((byte = message[i]) != 0) {
crc = (crc >> 8) ^ table[(crc ^ byte) & 0xFF];
i = i + 1;
}
return ~crc;
}
If you simply google "CRC32", you will get more info than you could possibly absorb.

In Memory QFile with fwrite

Rather than going to the filesystem to save and read back a PGM image file, I want to do this in memory. Can I somehow use a QBuffer as an in memory QFile to bypass saving to the filesystem:
QFile filename(QString("/home/pi/frame-%1.pgm").arg(i));
bool didOpen = filename.open(QIODevice::ReadWrite);
qDebug() << "Did open file: " << didOpen << filename.fileName();
int fileHandle = filename.handle();
FILE * f = fdopen(dup(fileHandle), "wb");
int res = fprintf(f, "P5 %d %d 65535\n", w, h);
for (int y = 0; y < h; y++) {
for (int x = 0; x < w; x++) {
uint16_t v = img[y*w+x];
//v = htobe16(v);
res = fwrite((uint8_t*)&v, sizeof(uint16_t), 1, f);
}
}
fclose(f);
QPixmap pixmap;
bool didLoad = pixmap.load(QString("/home/pi/frame-%1.pgm").arg(i));
emit updateScreen(pixmap);
Actually, yes.
You already have most of your data ready. We just need to turn it into a format that QPixmap can read directly. For that we use the QPixmap(const char *const[] xpm) constructor to make a pixmap from memory. By happy coincidence this constructor takes an array of pointers, not a straight array, which saves having to copy the bitmap data!
Untested code:
char *lines[] = (char **)malloc(sizeof(char *) * h + 1); // 1 extra for the header
char header[100];
sprintf(header, "P5 %d %d 65535\n", w, h);
lines[0] = header;
for (int y = 0; y < h; y++) {
lines[y + 1] = (char *)&img[y * w]; // note y+1 offset
}
QPixmap pixmap(lines);
emit updateScreen(pixmap);
free(lines);
Note: sizeof(char *) returns the size of a char pointer, so in the first line we allocate an array of pointers of h lines + 1 for the header. After setting the first "line" of the array to the header we copy the address offsets of your img memory block in the remaining "lines" and feed it to QPixmap. After that we're done.

UTF-8 to Latin(ISO-8859-1) Conversion in C++

I would like to know how to write code that performs a UTF-8 to Latin(ISO-8859-1) Conversion in C++.
The following website does the conversion required:
http://www.unicodetools.com/unicode/utf8-to-latin-converter.php
Inserting value: úsername
provides the result: úsername
I've got a piece of code that does a similar job from a previous post but doesn't seem to convert the string
int utf8_to_unicode(std::deque<int> &coded)
{
int charcode = 0;
int t = coded.front();
coded.pop_front();
if (t < 128)
{
return t;
}
int high_bit_mask = (1 << 6) -1;
int high_bit_shift = 0;
int total_bits = 0;
const int other_bits = 6;
while((t & 0xC0) == 0xC0)
{
t <<= 1;
t &= 0xff;
total_bits += 6;
high_bit_mask >>= 1;
high_bit_shift++;
charcode <<= other_bits;
charcode |= coded.front() & ((1 << other_bits)-1);
coded.pop_front();
}
charcode |= ((t >> high_bit_shift) & high_bit_mask) << total_bits;
return charcode;
}
Help please!
You need the iconv(3) function from libiconv. The first argument (some iconv_t) to the iconv conversion function should be obtained by iconv_open(3) at program initialization, probably with
ic = iconv_open("ISO-8859-1","UTF-8");
(where ic is some static or global iconv_t variable).

Thrust vectorized search: Efficiently combine lower_bound and binary_search to find both position and existence

I'm trying to use Thrust to detect if each element of an array can be found in another array and where (both arrays are sorted). I came across the vectorized search routines (lower_bound and binary_search).
lower_bound will return for each value the index where it could be inserted in a list respecting its ordering.
I also need to know if the value is found or not (which can be done with binary_search), not just its position.
Is it possible to achieve both efficiently without making two searches (calling binary_search and then lower_bound)?
I know in the scalar case, lower_bound will return a pointer to end of the array if a value cannot be found, but this does not happens in the vectorized version.
Thanks!
You can check that the element that lower_bound returns is the same as the one you searched for. E.g. given a = {1,3,5} and searching for b = {1,4}, the result will be c = {0,2}. We have a[c[0]] == b[0], so b[0] is in a, but a[c[1]] != b[1] so b[1] is not in a.
(Note that you will need to ensure that you don't make any out-of-bounds memory accesses, since lower_bound can return an index that is beyond the end of the array.)
#tat0: you can also play around with Arrayfire:
vectorized search using lower_bound() does not give you the answer immediately
while with setintersect() in arrayfire, you get the "intersection" of two arrays directly:
float A_host[] = {3,22,4,5,2,9,234,11,6,17,7,873,23,45,454};
int szA = sizeof(A_host) / sizeof(float);
float B_host[] = {345,5,55,6,7,8,19,2,63};
int szB = sizeof(B_host) / sizeof(float);
// initialize arrays from host data
array A(szA, 1, A_host);
array B(szB, 1, B_host);
array U = setintersect(A, B); // compute intersection of 2 arrays
int n_common = U.elements();
std::cout << "common: ";
print(U);
the output is:
common: U = 2.0000
5.0000
6.0000
7.0000
to get the actual locations of these elements in array A, you can use the following
construct (provided that elements in A are unique):
int n_common = U.elements();
array loc = zeros(n_common); // empty array
gfor(array i, n_common) // parallel for loop
loc(i) = sum((A == U(i))*seq(szA));
print(loc);
then: loc =
4.0000
3.0000
8.0000
10.0000
Furthermore, thrust::lower_bound() seems to be slower than setintersect(),
i benchmarked it with the following program:
int *g_data = 0;
int g_N = 0;
void thrust_test() {
thrust::device_ptr<int> A = thrust::device_pointer_cast((int *)g_data),
B = thrust::device_pointer_cast((int *)g_data + g_N);
thrust::device_vector<int> output(g_N);
thrust::lower_bound(A, A + g_N, B, B + g_N,
output.begin(),
thrust::less<int>());
std::cout << "thrust: " << output.size() << "\n";
}
void af_test()
{
array A(g_N, 1, g_data, afDevicePointer);
array B(g_N, 1, g_data + g_N, afDevicePointer);
array U = setintersect(A, B);
std::cout << "intersection sz: " << U.elements() << "\n";
}
int main()
{
g_N = 3e6; // 3M entries
thrust::host_vector< int > input(g_N*2);
for(int i = 0; i < g_N*2; i++) { // generate some input
if(i & 1)
input[i] = (i*i) % 1131;
else
input[i] = (i*i*i-1) % 1223 ;
}
thrust::device_vector< int > dev_input = input;
// sort the vector A
thrust::sort(dev_input.begin(), dev_input.begin() + g_N);
// sort the vector B
thrust::sort(dev_input.begin() + g_N, dev_input.begin() + g_N*2);
g_data = thrust::raw_pointer_cast(dev_input.data());
try {
info();
printf("thrust: %.5f seconds\n", timeit(thrust_test));
printf("af: %.5f seconds\n", timeit(af_test));
} catch (af::exception& e) {
fprintf(stderr, "%s\n", e.what());
}
return 0;
}
and the results:
CUDA toolkit 4.2, driver 295.59
GPU0 GeForce GT 650M, 2048 MB, Compute 3.0 (single,double)
Memory Usage: 1937 MB free (2048 MB total)
thrust: 0.13008 seconds
arrayfire: 0.06702 seconds

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