Reading a string with scanf

Question

I m a little bit confused about something  I was under the impression that the correct way of reading a C string with scanf   went along the lines of   never mind the possible buffer overflow  it s just a simple example   char string 256   scanf    s    string      However  the following seems to work too   scanf    s     amp string      Is this just my compiler  gcc   pure luck  or something else

User · Answer

I think that this below is accurate and it may help. Feel free to correct it if you find any errors. I'm new at C.

char str[]

array of values of type char, with its own address in memory
array of values of type char, with its own address in memory as many consecutive addresses as elements in the array
including termination null character '\0' &str, &str[0] and str, all three represent the same location in memory which is address of the first element of the array str

char *strPtr = &str[0]; //declaration and initialization

alternatively, you can split this in two:

char *strPtr; strPtr = &str[0];

strPtr is a pointer to a char
strPtr points at array str
strPtr is a variable with its own address in memory
strPtr is a variable that stores value of address &str[0]
strPtr own address in memory is different from the memory address that it stores (address of array in memory a.k.a &str[0])
&strPtr represents the address of strPtr itself

I think that you could declare a pointer to a pointer as:

char **vPtr = &strPtr;

declares and initializes with address of strPtr pointer

Alternatively you could split in two:

char **vPtr;
*vPtr = &strPtr

*vPtr points at strPtr pointer
*vPtr is a variable with its own address in memory
*vPtr is a variable that stores value of address &strPtr
final comment: you can not do str++, str address is a const, but you can do strPtr++

User · Answer

An array  decays  into a pointer to its first element  so scanf   s   string  is equivalent to scanf   s    amp string 0    On the other hand  scanf   s    amp 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  amp string 0   but when you ve passed in  amp string you re depending on something that the language standard doesn t guarantee  namely that the pointers  amp string and  amp 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

[c] Reading a string with scanf

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