unsigned-char

I have to call a win32 dll function int func1( int arg1, unsigned char **arg2, int *arg3); and i need wrapped in c# as public extern int fuc1(int arg1, out IntPtr arg2, out IntPtr arg3); and i called it from a c# application as int arg1; IntPtr arg2 = IntPtr.Zero; IntPtr arg3 = IntPtr.Zero; func1(arg1,out arg2,out arg3); Is the function declared in the c# wrapper as well as called in C# test app Correct ? Now i need to store the arg2 in a text file. How to do that. Got answer from Hans and i wrote it in a file using System.IO.StreamWriter(@Application.StartupPath + "\\Filename.txt"); file

I would like to print the following hashed data. How should I do it? unsigned char hashedChars[32]; SHA256((const unsigned char*)data.c_str(), data.length(), hashedChars); printf("hashedChars: %X\n", hashedChars); // doesn't seem to work?? The hex format specifier is expecting a single integer value but you're providing instead an array of char . What you need to do is print out the char values individually as hex values. printf("hashedChars: "); for (int i = 0; i < 32; i++) { printf("%x", hashedChars[i]; } printf("\n"); Since you are using C++ though you should consider using cout instead of

Apparently the compiler considers them to be unrelated types and hence reinterpret_cast is required. Why is this the rule? EdChum They are completely different types see standard: 3.9.1 Fundamental types [basic.fundamental] 1 Objects declared as characters char) shall be large enough to store any member of the implementation's basic character set. If a character from this set is stored in a character object, the integral value of that character object is equal to the value of the single character literal form of that character. It is implementation-defined whether a char object can hold

One often needs to read from memory one byte at a time, like in this naive memcpy() implementation: void *memcpy(void *dest, const void *src, size_t n) { char *from = (char *)src; char *to = (char *)dest; while(n--) *to++ = *from++; return dest; } However, I sometimes see people explicitly use unsigned char * instead of just char * . Of course, char and unsigned char may not be equal. But does it make a difference whether I use char * , signed char * , or unsigned char * when bytewise reading/writing memory? UPDATE: Actually, I'm fully aware that c=200 may have different values depending on

I have the following code: cvtColor (image, image, CV_BGRA2RGB); Vec3b bottomRGB; bottomRGB=image.at<Vec3b>(821,1232); When I display bottomRGB[0] , it displays a value greater than 255. What is the reason for this? As you have commented, the reason is that you use cout to print its content directly. Here I will try to explain to you why this will not work. cout << bottomRGB[0] << endl; Why "cout" works weird for "unsigned char" ? It will not work because here bottomRGB[0] is a unsigned char (with value 218 ), cout actually will print some garbage value (or nothing) as it is just a non

when the following code is compiled it goes into an infinite loop: int main() { unsigned char ch; FILE *fp; fp = fopen("abc","r"); if(fp==NULL) { printf("Unable to Open"); exit(1); } while((ch = fgetc(fp))!=EOF) printf("%c",ch); fclose(fp); printf("\n",ch); return 0; } The gcc Compiler also gives warning on compilation abc.c:13:warning: comparison is always true due to limited range of data type the code runs fine when unsigned char is replaced by char or int as expected i.e. it terminates. But the code also runs fine for unsigned int as well. as i have i have read in EOF is defines as -1 in