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British usage of “cha”, “char” or “chai” to mean “tea” By happenstance, I stumbled upon the words cha, char and chai in the dictionary today, all defined as meaning tea in informal British English I lived and worked in London for some time, but never
c++ - char and char* (pointer) - Stack Overflow For cout << q - operator << (ostream , char* p) expects that p points to NULL terminated string - and q points to memory containing "H" but what is after this character no one knows - so you will get some garbage on screen Use cout << q to print single character
Difference between char and char* in c - CS50 Stack Exchange The difference between char* the pointer and char[] the array is how you interact with them after you create them If you are just printing the two examples, it will perform exactly the same They both generate data in memory, {h, e, l, l, o, 0} The fundamental difference is that in one char* you are assigning it to a pointer, which is a variable In char[] you are assigning it to an array
c - char *array and char array [] - Stack Overflow char *array = "One good thing about music"; declares a pointer array and make it point to a (read-only) array of 27 characters, including the terminating null-character
How to pronounce the programmers abbreviation char In many programming languages, char is a type name for character values The word character is pronounced with a [k] sound, but what about char? While trying to find the answer elsewhere, I learn
c - Get a substring of a char* - Stack Overflow char *buff = "this is a test string"; printf("% *s", 4, buff + 10); You could achieve the same thing by copying the substring to another memory destination, but it's not reasonable since you already have it in memory This is a good example of avoiding unnecessary copying by using pointers
Error: format specifies type char - CS50 Stack Exchange The problem is that you are taking the address of a pointer The way you have declared name it's type is effectively of a pointer to an array of 25 chars And thus name is a pointer to a pointer You just need to remove the in front of name in the scanf statement Strongly suggest you go back and watch the lesson again Also a simple google search for "char array scanf" would have yielded
c - Is it possible to convert char - Stack Overflow It sounds like you're confused between pointers and arrays Pointers and arrays (in this case char * and char []) are not the same thing An array char a[SIZE] says that the value at the location of a is an array of length SIZE A pointer char *a; says that the value at the location of a is a pointer to a char This can be combined with pointer arithmetic to behave like an array (eg, a[10] is