There is one semicolon after the if statement that is causing the problem. if (i!=0); { //this will always execute } change it to if (i!=0) { //this will execute if i != 0 } The compiler does not warn you because the first statement is syntactically valid.
The pre-conditions of the printf function require that you pass a data pointer as an argument to %p – and more specifically, a pointer of type void*. obj1 is not a data pointer; it is an instance of issue. The behaviour of violating the pre-conditions of printf is undefined. obj1.issue_c2 is a pointer, but it … Read more
From your tags, I’m guessing you’re using printf(). Simply leave out the “\n” character, which means “newline.”. In other words. printf(“> “); printf(“text\n”); This will print: > text