What is the problem of this extremely small C program?

incorporate
#include
#include

int main(int argc, char *argv)

int post = 0;
while(i! = 5);

printf(“%i\n”, i);
i++;

No do whatever!!!

At the chance of delivering the answer to a quiz, I’ll issue you on this line of code:

while(i! = 5);

There’s a skillful mistake upon it.Review it CAREFULLY.Hint:Because of this mistake, the code in the while-loop will never execute.

Expect that helps.

It’s to create accidentally set a semicolon “; ” once the while(i! =5)
POST deleted which colon, and now the when should trigger the stuff within the brackets.

#include
#include
#include

int main(int argc, char *argv)

int post = 0;
while(i! = 5) //<–fixed this

printf(“%i\n”, i);
i++;

I think your dilemma is with the int main(int argc, char *argv) line because by what post remember args is often a ‘reserved’ word in C.In your code I do think the compiler doesnt know very well what the argc and also argv things are so it cant can anything with them.
Do you get errors when you compile it
I believe your code must be the adhering to:

#include
#include
#include

int main(int argc, char *args)

int post = 0;
while(i! = 5);

printf(“%i\n”, i);
i++;

change to

incorporate
#include
#include

int main(int argc, char *argv)

int post = 0;
while(i! = 5);

printf(“i\n”, $=% i);
i++;