C program, strlen not counting whitespace?

Question

i've written a simple program, where the program takes and imput and prints out the characters in sequence and counts them

char name[100];
int counter=0
printf("Enter name and surname:");
scanf("%s",&name);
    
for(int i=0; i<(strlen(name)); i++){
        printf("%c \n",name[i]);
        counter++;
printf("%d",counter)

for some reason , when i imput John Snow, it prints out only

J
o
h
n 
4

desired output

J
o
h
n

S
n
o
w
10

any help please? within reasonable skill level :D if posible no malloc() free() and memory things, its supposed to be quite easy :D

BTW, get rid of `&` before `name`. Arrays automatically decay to pointers when used as function arguments. — Barmar, Nov 16 '21 at 17:24
You could read a whole line instead, using e.g. `fgets`. But then you might want to read [Removing trailing newline character from fgets() input](https://stackoverflow.com/questions/2693776/removing-trailing-newline-character-from-fgets-input) as well. — Some programmer dude, Nov 16 '21 at 17:25
Alternatively, you can use [`getline()`](https://man7.org/linux/man-pages/man3/getline.3.html) instead — Remy Lebeau, Nov 16 '21 at 17:25
Just remember to remove the newline at the end. See https://stackoverflow.com/questions/2693776/removing-trailing-newline-character-from-fgets-input — Barmar, Nov 16 '21 at 17:25
BTW, `strlen()` does count white-spaces. It is just that `name[]` does not contain any white-space. — chux - Reinstate Monica, Nov 16 '21 at 18:05

Vlad from Moscow · Answer 1 · 2021-11-16T17:39:47.627

1

For starters you have to write

scanf("%s",name);

instead of

scanf("%s",&name);

Secondly when the format string "%s" is used the function stops to enter characters as soon as a white space character is encountered.

You should write instead

scanf( " %99[^\n]", name );

Another approach is to use the function fgets

fgets( name, sizeof( name ), stdin );

The function can append the new line character '\n' to the entered string. To remove it you can write

name[ strcspn( name, "\n" ) ] = '\0';

edited Nov 16 '21 at 17:39

answered Nov 16 '21 at 17:28

Vlad from Moscow

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when i put scanf( "%99[^\n]", name ); it automaticly skips imput, doesnt let me enter anything – D. K. Nov 16 '21 at 17:35
2

@D.K. See the blank before the symbol % in the format string. – Vlad from Moscow Nov 16 '21 at 17:40

C program, strlen not counting whitespace?

1 Answers1