I'm writing some C code to parse IEEE 802.11 frames, but I'm stuck trying to create a new variable whose length depends on the size of the frame itself.
Here's the code ... |
I have declared a char variable ope in function main.
i took the input through getchar function and stored in ope.
Now i want to blank the variable so i will be able ... |
I think my problem is something simple, but I'm not seeing it. I'm new to programming in C and this is an effort to see what I've absorbed, bit by bit. ... |
Several things here: 1) Functions like strlen() work with char array that contain C-stirngs. That is, arrays where the character following the last character of the string is a binary zero. You have declared a char array but there is nothin to indicate that there is a C-string in v_deal_client. If there isn't, then strlen won't work. 2) 16-v_deal_length is a ... |
problem in C++ -variable of type std::string called str egl std:string str; // declaration str+ ""; // initialized to this.........nothing. -we need to pass the value of this variable to a function which requires a variable of type char *. egl drawString(screen, font, o,o, str) ? -the easiest thing to do is convert the str variable to a char * type ... |
owolablo wrote: Can anybody please tell me how to change the individual elements of a char variable. I need to parse through the string, check for a particular character and change it to something else if it is found. Then I guess you don't mean a char variable, but rather an array of char, otherwise there is only one single character ... |
I'm using the gcc with the -pedantic setting. On my embedded system I need a parameter class that contains all setting for the software. This class needs to be ROM-able to locate it in some kind of eprom memory. Thus, all members of this class are const and there is only one static instance. An example header: class MyParameters { public: ... |
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#include int main ( void ) { char left[256] = {0}; char name[4]; char right[256] = {0}; printf ( "Enter your name: " ); fflush ( stdout ); scanf ( "%s", name ); printf ( "left: |%s|\n", left ); printf ( "name: |%s|\n", name ); printf ( "right: |%s|\n", right ); return 0; } |
Hello friends, I am making a simple to calculate the digits of a number. The basic code is fine and it is working. Now I want that if the user is interested he can input another no and found the sum of digit for it. For this I am using a while loop with condition that a char variable x has ... |
Code: #include #include #include int main() { int inLoop = 0, outLoop = 0, input; int numberOne, numberTwo; char inputChar; while (outLoop == 0) { srand( time(NULL) ); while (inLoop == 0) { numberOne = rand(); numberTwo = rand(); printf("What is %d times %d? ", numberOne, numberTwo); scanf("%d", &input); if (input == (numberOne * numberTwo) ) { printf("Correct!\n"); ... |
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#include int main(void) { char chInput; while(chInput != 'E') { //while the user dont hit E printf("Enter a char: E for exit!\n"); chInput = getchar(); fflush(stdin); if((chInput == 'A') || (chInput == 'a')) printf("You did hit the A letter\n"); else printf("You did hit a letter that its not A\n"); } return 0; } |
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__________________ Right 98% of the time, and don't care about the other 3%. It has been said that the great scientific disciplines are examples of giants standing on the shoulders of other giants. It has also been said that the software industry is an example of midgets standing on the toes of other midgets. (Alan Cooper) |
Expecting the language to be able to do that is very wishful thinking! If you want to evaluated user entered expressions you need to provide your own expression evaluator. You need ot interprete user input, but C is compiled language, it has no capability to interpret runtime input. A simple solution to the problem you posted is possible, as described by ... |
The above poster is not exactly correct. When you call scanf(), if the function fails (which will happen when you try to read an integer but it is given a character), it will not actually consume the input character from the input. This means the next time you call scanf() it will end up trying to read the same character again! ... |
#include #define INT_DIGIT 21 /* Supports 64-bit int */ int main ( void ) { char b2b_data_Out[INT_DIGIT]; int x; printf ( "Enter a value: " ); fflush ( stdout ); if ( scanf ( "%d", &x ) == 1 ) { sprintf ( b2b_data_Out, "%d", x ); puts ( b2b_data_Out ); } return 0; } |