[Solved] problems with scanf and conversion specifiers

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These are the very basics of C Programming, and I strongly advise you to get a decent book – The C Programming Language by Dennis Ritchie would be a good start.

There are numerous errors in your code.

  1. A char can contain only one character, like ‘A’, or ‘a’ or something like that. When you’re scanning a name, it is going to be a group of characters, like ‘E’, ‘d’, ‘d’, ‘y’. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.

  2. When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c\n", charVar) is sufficient.

  3. scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input – and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x – which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn’t need to be preceded by an ampersand.

I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. 🙂

solved problems with scanf and conversion specifiers