Input:9 notation
printf("Input N value\n");
scanf("%d",&N);
char X[N];
int i=0,N=0;
for (i=0;i<N-1;i++)
{
scanf("%c",&X[i]);
}
X[N-1]='\0';
for (i=0;i<N-1;i++)
{
printf("%c",X[i]);
}
Expected output:notation Output:notatio
Why is this so?
-
corrected my answer! (Don't know why i missed that one)Rizier123– Rizier1232015-01-01 11:29:19 +00:00Commented Jan 1, 2015 at 11:29
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2 Answers
When you read the length with your first scanf call you leave a newline in your input buffer.
At the second scanf call the first character read is the newline and then the rest. Adding a space before the %c in your second scanf string parameter will consume any left newlines.
scanf(" %c",&X[i]);
You also don't null terminate your string, before printing it.
X[N-1] = '\0' ;
Comments
You dont have to substract 1 in your for loops, otherwise it will miss 1 character at the end!
This should work:
#include <stdio.h>
int main() {
int charNumber, charCount;
printf("Input N value\n>");
scanf("%d", &charNumber);
char text[charNumber+1];
for (charCount = 0; charCount < charNumber; charCount++)
scanf(" %c", &text[charCount]);
text[charNumber] = '\0';
for (charCount = 0; charCount <= charNumber; charCount++)
printf("%c", text[charCount]);
return 0;
}
You can also print the string with:
printf("%s", text);
