The following code is completely ok in C No, Not at all. In your code if(ch==”a”) is essentially trying to compare the value of ch with the base address of the string literal “a”,. This is meaning-and-use-less. What you want here, is to use single quotes (‘) to denote a char literal, like if(ch == … Read more
You can’t assign anything to an array variable in C. It’s not a ‘modifiable lvalue’. From the spec, ยง6.3.2.1 Lvalues, arrays, and function designators: An lvalue is an expression with an object type or an incomplete type other than void; if an lvalue does not designate an object when it is evaluated, the behavior is … Read more
In fact, you can do so using a compound literal, a feature added to the language by the 1999 ISO C standard. A string literal is of type char[N], where N is the length of the string plus 1. Like any array expression, it’s implicitly converted, in most but not all contexts, to a pointer … Read more
Interesting. I had more luck sending IAC WILL ECHO IAC WILL SUPPRESS_GO_AHEAD IAC WONT LINEMODE 255 251 1 255 251 3 255 252 34 The IAC WONT LINEMODE seems to be redundant: my telnet client seems to get to the right state without it, but I left it in for completeness.
C++ has something called “multicharacter literals”. ‘1234’ is an example of one. They have type int, and it is implementation-defined what value they have and how many characters they can contain. It’s nothing directly to do with the fact that characters are represented as integers, but chances are good that in your implementation the value … Read more
C does not and never has had a native string type. By convention, the language uses arrays of char terminated with a null char, i.e., with ‘\0′. Functions and macros in the language’s standard libraries provide support for the null-terminated character arrays, e.g., strlen iterates over an array of char until it encounters a ‘\0’ … Read more
This might help, I used to fi my files like this: http://security102.blogspot.ru/2010/04/findreplace-of-nul-objects-in-notepad.html Basically you need to replace \x00 characters with regular expressions
The easiest way to for-each every char in a String is to use toCharArray(): for (char ch: “xyz”.toCharArray()) { } This gives you the conciseness of for-each construct, but unfortunately String (which is immutable) must perform a defensive copy to generate the char[] (which is mutable), so there is some cost penalty. From the documentation: … Read more
You can’t fit 5 bytes worth of data into a 4 byte array; that leads to buffer overflows. If you have the hex digits in a string, you can use sscanf() and a loop: #include <stdio.h> #include <ctype.h> int main() { const char *src = “https://stackoverflow.com/questions/1557400/0011223344”; char buffer[5]; char *dst = buffer; char *end = … Read more
// Yet, another good itoa implementation // returns: the length of the number string int itoa(int value, char *sp, int radix) { char tmp[16];// be careful with the length of the buffer char *tp = tmp; int i; unsigned v; int sign = (radix == 10 && value < 0); if (sign) v = -value; … Read more