c - Why does the presence of `n` in the input sentence before `\n` give wrong output? -


i want input sentence (containing possible characters) , print it. there catch. if there \n in sentence part of sentence before \n should printed out (i.e. \n should signify end of inputted sentence). wrote code situation : 

#include <stdio.h> main() {     char ch[100];     printf("enter sentence");     scanf("%99[^\\n]",&ch);     printf("%s",ch); }

this code seems work fine fails in situation. if there character n anywhere in sentence before \n prints first word of sentence! why happen? how can fix bug?

this case works fine: enter image description here

but in case fails:

enter image description here

detail comments:

q: want to stop @ newline, or @ backslash followed n?
a: slash followed n

the [] conversion specifier of scanf() works defining accepted (or, ^, rejected) set of characters. %[^\\n] stop scanning @ first \ or first n -> can't solve problem scanf().

you should read line of input fgets() , search occurence of "\\n" strstr().

side note: there's error in program:

char ch[100]; scanf("%99[^\\n]",&ch);

ch evaluates pointer first element of array (so, fine parameter scanf()), while &ch evaluates pointer array, not scanf() expects.

(the difference in type, address same)


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