Why not use ampersent

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#include <stdio.h>
    int main(){ 

    char st[55];


    printf("Enter name \n");
    scanf("%c\n",st);<pre>// why we don'nt need to add & in it

return 0;
}

What I have tried:

I tried to take user from input without & in a string

Posted 9-Jun-22 0:36am

Aayush Jain 2022

Updated 9-Jun-22 1:33am

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4 solutions

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Greg Utas · Accepted Answer · 2022-06-09T00:55:00

Solution 1

When you pass an array as an argument in C or C++, it is treated as a pointer to the beginning of the array, so passing st here is basically the same as passing &st[0].

Posted 9-Jun-22 0:55am

Greg Utas

Patrice T · Accepted Answer · 2022-06-09T00:59:00

Solution 2

Quote:
Why not use ampersent

Because the compiler understand what you want to do, and it does the necessary adjustments automatically.
The compiler understands that you want to use a pointer,
scanf - C++ Reference[^]
C library function - scanf()[^]

Posted 9-Jun-22 0:59am

Patrice T

OriginalGriff · Accepted Answer · 2022-06-09T01:28:00

Solution 3

The C language specification says that the name of an array is a pointer to the first element of that array - so passing st is the equivelant of typing &(st[0]) but a lot easier to read!

Posted 9-Jun-22 1:28am

OriginalGriff

Updated 9-Jun-22 1:29am

v2

Richard MacCutchan · Accepted Answer · 2022-06-09T01:33:00

Solution 4

Get yourself a copy of The C Programming Language - Google Search[^] where all such issues are clearly explained.

As suggested last week.

Posted 9-Jun-22 1:33am

Richard MacCutchan

Updated 9-Jun-22 1:35am

v2