String input using C scanf_s

From the specification of fscanf_s() in Annex K.3.5.3.2 of the ISO/IEC 9899:2011 standard:

The fscanf_s function is equivalent to fscanf except that the c, s, and [ conversion specifiers apply to a pair of arguments (unless assignment suppression is indicated by a *). The first of these arguments is the same as for fscanf. That argument is immediately followed in the argument list by the second argument, which has type rsize_t and gives the number of elements in the array pointed to by the first argument of the pair. If the first argument points to a scalar object, it is considered to be an array of one element.

and:

The scanf_s function is equivalent to fscanf_s with the argument stdin interposed before the arguments to scanf_s.

MSDN says similar things (scanf_s() and fscanf_s()).

Your code doesn't provide the length argument, so some other number is used. It isn't determinate what value it finds, so you get eccentric behaviour from the code. You need something more like this, where the newline helps ensure that the output is actually seen.

char name[64];
if (scanf_s("%s", name, sizeof(name)) == 1)
    printf("Your name is %s\n", name);

I used this very often in my university classes so this should work fine in Visual Studio (tested in VS2013):

char name[64]; // the null-terminated string to be read
scanf_s("%63s", name, 64);
// 63 = the max number of symbols EXCLUDING '\0'
// 64 = the size of the string; you can also use _countof(name) instead of that number
// calling scanf_s() that way will read up to 63 symbols (even if you write more) from the console and it will automatically set name[63] = '\0'
// if the number of the actually read symbols is < 63 then '\0' will be stored in the next free position in the string
// Please note that unlike gets(), scanf() stops reading when it reaches ' ' (interval, spacebar key) not just newline terminator (the enter key)
// Also consider calling "fflush(stdin);" before the (eventual) next scanf()

Ref: https://msdn.microsoft.com/en-us/library/w40768et.aspx

Here is a part of code that works for me fine:

char name[64];
scanf_s("%63s", name,(unsigned)_countof(name));
printf("Your name is %s", name);

For more information here is a link: https://learn.microsoft.com/de-de/cpp/c-runtime-library/reference/scanf-s-scanf-s-l-wscanf-s-wscanf-s-l?view=msvc-170

Best Regards

The scanf_s function is equivalent to scanf except that %c, %s, and %[ conversion specifiers each expect two arguments (the usual pointer and a value of type rsize_t indicating the size of the receiving array, which may be 1 when reading with a %c into a single char)

Your code doesn't provide the size of receiving array, also the variable name is a pointer pointing to the first character of the array, so it contains the address of name[0]. Therefore your first argument name in scanf_s is correct because name is a pointer, also note that, for the second argument you can't insert the size of a pointer like sizeof(name) because it is always same. You need to specify the size of your char array (name[64]), so for the second argument you should insert sizeof(name) or sizeof(char[64]) or 64*sizeof(char).

You can correct your code as follows:

char name[64];
if (scanf_s("%s", name, sizeof(name)) == 1)
    printf("Your name is %s\n", name);

you should do this : scanf ("%63s", name);

Update:

The below code worked for me:

#include <stdio.h> 
int main(void) { 
    char name[64]; 
    scanf ("%63s", name); 
    printf("Your name is %s", name); 
    return 0; 
}

if you are using visual studio, go to Project properties -> Configuration Properties -> C/C++-> Preprocessor -> Preprocessor Definitions click on edit and add _CRT_SECURE_NO_WARNINGS click ok, apply the settings and run again.

Note: this is only good if you are doing your homework or something like that and it's not recommended for production.

    #include<stdio.h>

    int main()
    {
      char name[64];

      printf("Enter your name: ");
      scanf("%s", name);
      printf("Your name is %s\n", name);

     return 0;
    }