I'm trying to run the code demo for ICU4C bellow, and getting
warning: implicit declaration of function 'austrdup'
which subsequently generate an error. I understand that this is due to the missing imported library that contains 'austrdup' function, and have been looking at the source code to guess which one it is, but no luck. Does anyone have any idea which one should be imported?
#include <unicode/umsg.h>
#include <unicode/ustring.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char const *argv[])
{
UChar* str;
UErrorCode status = U_ZERO_ERROR;
UChar *result = NULL;
UChar pattern[100];
int32_t resultlength, resultLengthOut, i;
double testArgs[] = { 100.0, 1.0, 0.0};
str=(UChar*)malloc(sizeof(UChar) * 10);
u_uastrcpy(str, "MyDisk");
u_uastrcpy(pattern, "The disk {1} contains {0,choice,0#no files|1#one file|1<{0,number,integer} files}");
for(i=0; i<3; i++){
resultlength=0;
resultLengthOut=u_formatMessage( "en_US", pattern, u_strlen(pattern), NULL, resultlength, &status, testArgs[i], str);
if(status==U_BUFFER_OVERFLOW_ERROR){ //check if output truncated
status=U_ZERO_ERROR;
resultlength=resultLengthOut+1;
result=(UChar*)malloc(sizeof(UChar) * resultlength);
u_formatMessage( "en_US", pattern, u_strlen(pattern), result, resultlength, &status, testArgs[i], str);
}
printf("%s\n", austrdup(result) ); //austrdup( a function used to convert UChar* to char*)
free(result);
}
return 0;
}
austrdup is not an official ICU method. It's only used by tests in ICU and defined in icu4c/source/test/cintltst/cintltst.h and implemented in icu4c/source/test/cintltst/cintltst.c. It is bascially just a wrapper around u_austrcpy.
Related
Since I just created this crude test using the functions from IETF RFC 4634, I don't know for certain whether I've used them correctly for HMAC-SHA-384-192, so I'll start with that code here:
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include "sha.h"
int main(int argc, char *argv[]) {
HMACContext hmac;
const unsigned char *keyarr = "0123456789abcdef0123456789abcdef0123456789abcdef0123456789abcdef";
int err = hmacReset(&hmac, SHA384, keyarr, 48);
if (err != shaSuccess) {
printf("err 1\n");
exit(1);
}
const uint8_t testarray[65] = {'I',' ','a','m',' ','n','o','t',' ','a',' ','c','r','o','o','k','!'};
const unsigned char *prfkey = "abcdef0123456789abcdef0123456789abcdef0123456789abcdef0123456789";
memcpy((void *)testarray + 17, (void *)prfkey, 48);
const int testlen = 65;
err = hmacInput(&hmac, testarray, testlen);
if (err != shaSuccess) {
printf("err 2\n");
exit(1);
}
uint8_t Message_Digest[USHAMaxHashSize];
err = hmacResult(&hmac, Message_Digest);
if (err != shaSuccess) {
printf("err 3\n");
exit(1);
}
int i;
for(i = 0; i < 24; i++) printf(" %02X", Message_Digest[i]);
putchar('\n');
}
If I've done everything right (other than selecting good keys) so far, I would ordinarily have a 24-byte (i.e., 192-bit) digest, but if I digitally sign the digest prior to appending it, my experience is that the signature block isn't a predictable length. I'm sure I could come up with any number of ways to identify the end of the message portion, but I don't want to make this a hack. What is the accepted way of doing this? (The signature will use ECDSA.)
I should also mention that this will be a multicast message using UDP inside ESP, since that puts constraints on message economy. (That's also the main reason for the problem--keeping it binary. The other is the practice of appending, rather than prefixing it with a byte count in front of it.)
I know that it might be a stupid question but can you tell me why the following patch of code fails? I see nothing wrong. I am trying to read integers using scanf. I have included the necessary library, but when I run the program it crashes after I read the first s. Thank you.
#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <vector>
using namespace std;
int main()
{
int n, x;
scanf("%d", &n); scanf("%d", &x);
vector< pair<int, int> > moments;
for(int i = 0; i < n; ++i)
{
int f, s;
scanf("%d", &f);
scanf("%d", &s );
moments[i].first = f;
moments[i].second = s;
}
return 0;
}
That is not the way to assign values to moments since moments[i] does not yet exist. Try:
pair<int, int> thing;
thing = make_pair(f,s);
moments.push_back(thing);
instead of your assignements to moments elements.
How can I modify the source code in the func( ) so that the address to which the program returns after executing func () is changed in such a manner that the instruction printf("first print\n”) is skipped. Use the pointer *ret defined in func() to modify the return address appropriately in order to achieve this.
Here is the code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void func(char *str)
{
char buffer[24];
int *ret;
strcpy(buffer,str);
}
int main(int argc, char **argv)
{
if (argc < 2)
{
printf("One argument needed.\n");
exit(0);
}
int x;
x = 0;
func(argv[1]);
x = 1;
printf("first print\n");printf("second print\n");
}
As sherrellbc noted, a program's exploits are usually written without modifying its source code. But if you want, inserting these two lines into func() may do:
ret = (int *)&str; // point behind saved return address
ret[-1] += 12; // or however many code bytes are to be skipped
I am new to CUDA and I am getting a strange error. I want to print a string from a passed object and I get the error "calling host function from global function is not allowed" and I don't know why. But if I want to print an integer (changing get method to return sk1), everything works fine. Here is the code:
class Duomenys {
private:
string simb;
int sk1;
double sk2;
public:
__device__ __host__ Duomenys(void): simb(""), sk1(0), sk2(0.0) {}
__device__ __host__~Duomenys() {}
__device__ __host__ Duomenys::Duomenys(string simb1, int sk11, double sk21)
: simb(simb1), sk1(sk11), sk2(sk21) {}
__device__ __host__ string Duomenys::get(){
return simb;
}
};
And here I am calling Duomenys::get from __global__ function:
__global__ void Vec_add(Duomenys a) {
printf(" %s \n",a.get());
}
EDIT: I am trying to read data from a file and print it in a global function. In this code I am trying read all data and print just one object to see if everything works. This is the error I'm getting:
calling a __host__ function("std::basic_string<char, std::char_traits<char>, std::allocator<char> >::~basic_string") from a __global__ function("Vec_add") is not allowed
Code:
#include <stdio.h>
#include <stdlib.h>
#include <cuda.h>
#include <cuda_runtime.h>
#include <vector>
#include <string>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
#include <sstream>
using namespace std;
class Duomenys {
private:
string simb;
int sk1;
double sk2;
public:
__device__ __host__ Duomenys(void): simb(""), sk1(0), sk2(0.0) {}
__device__ __host__~Duomenys() {}
__device__ __host__ Duomenys::Duomenys(string simb1, int sk11, double sk21)
: simb(simb1), sk1(sk11), sk2(sk21) {}
__device__ __host__ string Duomenys::print()
{
stringstream ss;
ss << left << setw(10) << simb << setw(10) << sk1 << setw(10) << sk2;
return ss.str();
}
};
__global__ void Vec_add(Duomenys a) {
printf(" %s \n",a.print());
}
/* Host code */
int main(int argc, char* argv[]) {
setlocale (LC_ALL,"");
vector<Duomenys> vienas;
vector<vector<Duomenys>> visi;
//data reading to vector "vienas" (it works without any errors)
Duomenys *darr;
const size_t sz = size_t(2) * sizeof(Duomenys);
cudaMalloc((void**)&darr, sz);
Vec_add<<<1, 1>>>(visi[0].at(0));
cudaDeviceSynchronize();
cudaMemcpy(darr, &visi[0].at(0), sz, cudaMemcpyHostToDevice);
return 0;
}
Your problem is not with printf function, but with string data type. You cannot use the C++ string type in a kernel. See related question here: Can we use the string data type in C++ within kernels
Why would you pass a string object to printf when the %s format specifier is expecting something else? When I try to do that in ordinary host code, I get warnings about "passing non-POD types through ellipsis (call will abort at runtime)". Note that this problem has nothing to do with CUDA.
But beyond that issue, presumably you're getting string from the C++ standard library. (It's better if you show a complete reproducer code, then I don't have to guess at where you're getting things or what you are including.)
If I get string as follows:
#include <string>
using namespace std;
Then I am using a function defined in the C++ Standard Library. CUDA supports the C++ language (mostly) but does not necessarily support usage of C++ libraries (or C libraries, for that matter) in device code. Libraries are (usually) composed of (at least some) compiled code (such as allocators, in this case), and this code has been compiled for CPUs, not for the GPU. When you try to use such a CPU compiled routine (e.g. an allocator associated with the string class) in device code, the compiler will bark at you. If you include the complete error message in the question, it will be more obvious specifically what (compiled-for-the-host) function is actually the issue.
Use a standard C style string instead (i.e. char[] and you will be able to use it directly in printf.
EDIT: In response to a question in the comments, here is a modified version of the code posted that demonstrates how to use an ordinary C-style string (i.e. char[]) and print from it in device code.
#include <stdio.h>
#include <stdlib.h>
#include <cuda.h>
#include <cuda_runtime.h>
#include <vector>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
#include <sstream>
#define STRSZ 32
using namespace std;
class Duomenys {
private:
char simb[STRSZ];
int sk1;
double sk2;
public:
__device__ __host__ Duomenys(void): sk1(0), sk2(0.0) {}
__device__ __host__~Duomenys() {}
__device__ __host__ Duomenys(char *simb1, int sk11, double sk21)
: sk1(sk11), sk2(sk21) {}
__device__ __host__ char * print()
{
return simb;
}
__device__ __host__ void store_str(const char *str)
{
for (int i=0; i< STRSZ; i++)
simb[i] = str[i];
}
};
__global__ void Vec_add(Duomenys a) {
printf(" %s \n",a.print());
}
/* Host code */
int main(int argc, char* argv[]) {
string host_string("hello\n");
setlocale (LC_ALL,"");
vector<Duomenys> vienas(3);
vienas[0].store_str(host_string.c_str());
vector<vector<Duomenys> > visi(3);
visi[0] = vienas;
//data reading to vector "vienas" (it works without any errors)
Duomenys *darr;
const size_t sz = size_t(2) * sizeof(Duomenys);
cudaMalloc((void**)&darr, sz);
Vec_add<<<1, 1>>>(visi[0].at(0));
cudaDeviceSynchronize();
cudaMemcpy(darr, &(visi[0].at(0)), sz, cudaMemcpyHostToDevice);
return 0;
}
Note that I didn't try to understand your code or fix everything that looked strange to me. However this should demonstrate one possible approach.
The question is in the title. Need help figuring out why my code compiles but doesn't work as intended. Thanks!
//This example demonstrates how to do vector<string> to vectro<int> conversion using a function.
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
using namespace std;
vector<int>* convertStringVectorToIntVector (vector<string> *vectorOfStrings)
{
vector<int> *vectorOfIntegers = new vector<int>;
int x;
for (int i=0; i<vectorOfStrings->size(); i++)
{
stringstream str(vectorOfStrings->at(i));
str >> x;
vectorOfIntegers->push_back(x);
}
return vectorOfIntegers;
}
int main(int argc, char* argv[]) {
//Initialize test vector to use for conversion
vector<string> *vectorOfStringTypes = new vector<string>();
vectorOfStringTypes->push_back("1");
vectorOfStringTypes->push_back("10");
vectorOfStringTypes->push_back("100");
delete vectorOfStringTypes;
//Initialize target vector to store conversion result
vector<int> *vectorOfIntTypes;
vectorOfIntTypes = convertStringVectorToIntVector(vectorOfStringTypes);
//Test if conversion is successful and the new vector is open for manipulation
int sum = 0;
for (int i=0; i<vectorOfIntTypes->size(); i++)
{
sum+=vectorOfIntTypes->at(i);
cout<<sum<<endl;
}
delete vectorOfIntTypes;
cin.get();
return 0;
}
The code above has only one problem: You are deleting your vectorOfStringTypes before you pass it to your conversion function.
Move the line delete vectorOfStringTypes; to after you have called your convert function and the program works as intended.