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We can take string input in C using scanf(“%s”, str).
1) Read string with spaces by using "%[^\n]" format specifier. The format specifier "%[^\n]" tells to the compiler that read the characters until "\n" is not found. See the output, now program is able to read complete string with white space.
In C scanf() requires pointers to the variable in which the value is to be stored. Therefore we pass address of the variables using & where as the string variable stores the address of the first location of the string. Hence we dont need to use & while reading strings.
In scanf() you usually pass an array to match a %s specifier, then a pointer to the first element of the array is used in it's place. For other specifiers like %d you need to pass the address of the target variable to allow scanf() to store the result in it.
An array "decays" into a pointer to its first element, so scanf("%s", string) is equivalent to scanf("%s", &string[0]). On the other hand, scanf("%s", &string) passes a pointer-to-char[256], but it points to the same place.
Then scanf, when processing the tail of its argument list, will try to pull out a char *. That's the Right Thing when you've passed in string or &string[0], but when you've passed in &string you're depending on something that the language standard doesn't guarantee, namely that the pointers &string and &string[0] -- pointers to objects of different types and sizes that start at the same place -- are represented the same way.
I don't believe I've ever encountered a system on which that doesn't work, and in practice you're probably safe. None the less, it's wrong, and it could fail on some platforms. (Hypothetical example: a "debugging" implementation that includes type information with every pointer. I think the C implementation on the Symbolics "Lisp Machines" did something like this.)
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