A linked list cannot be represented by an array like you are trying to do, it is a different data structure. In the simplest form it is a structure containing a value and a pointer to the next...
Type: Posts; User: iceaway
A linked list cannot be represented by an array like you are trying to do, it is a different data structure. In the simplest form it is a structure containing a value and a pointer to the next...
You can use two nested loops, where the outer one loops through the rows and the inner one the columns, where the counter variables represent the current row and column. Something like:
int i,...
Show us the code you have so far.
What sort of programming are you interested in? Low level stuff? Kernel development? GUI applications?
You must know the size of your array. You don't necessarily have to know how large it is going to be at compile time, it can be allocated and resized dynamically at run time using malloc/realloc, but...
No. Drop the #define at the top, instead write
/* This is the function definition, you need this if the function is defined after it is used */
void Simulation(void);
int main(void) /*...
It's called "calling a function".
Oh you mean the name of '}'? :) Curly bracket, or curly brace, I think.
Link
You seem to be calling Simulation from your main loop but the function definition says it's called simulaition. C is case sensitive, and the spelling is wrong in the definition.
You should (almost) never use & with printf, as it takes the value of its arguments, not a pointer (except when using strings...). You should (almost) always use & with scanf as it need to write to...
You need to open a new file for writing using fopen, then you write to it using fprintf (almost identical syntax to printf) and close it again with fclose. Or you can do what Salem says and redirect...
Make it a tab or comma separated file, you can import that to excel.
The string you are writing to s contains 6 elements (remember the terminating null character), but you only allocated memory for 5.
Do you have any code at all to start with? You could at least write a skeleton which has a main function and the string-replace function.
You should close the FILE pointers, not the file names. i.e., fclose(dataIn) rather than fclose(dataInName). You cannot "close" a string :-)
What are you having problems with?
I'm still getting this error:
warning: format '%e' expects type 'float *', but argument 2 has type 'double *'
The format specifier for a double in scanf is %lf, and for printf it is %f. Try...
I rarely do anything with strings, so I'm not the person to answer that question. Hopefully CommonTater will come around soon, he should have some advice on this matter.
It is because you are trying to put characters which aren't in the ASCII-table into a char. Encoding is a headache, look into things like UTF-8, unicode etc. if you want to know more about it. The...
You already know what a and b is at the end of function2, 5 and 2. function2 returns 2*a - b, which would be 10 - 2 = 8. The return value from function2 is assigned to variable b in main. Since...
Try to do what I just did. Comment the code, line by line (or do it on paper) to get a feeling of what happens. The second line to be printed is the call to printf just after the call to function2()....
I think this belongs in the C++ forum.
I commented the code below for you:
int function2(int a, int b) { /* called with a = 1, b = 2 */
int temp = a+b; /* temp = a + b = 1 + 2 = 3 */
...
The first thing that will be printed is the printf-statement in function2(...). You know the parameter values passed to function2, so just follow the statements in each row, writing down the contents...
When the string literal "%d" is used as an argument to printf, it becomes a pointer to the address of the string. So when +1 is added to it, the pointer adress is incremented by one and points to "d"...