Simple C scanf does not work?

Question

If I try something such as:

int anint;
char achar;

printf("\nEnter any integer:");
scanf("%d", &anint);
printf("\nEnter any character:");
scanf("%c", &achar);
printf("\nHello\n");
printf("\nThe integer entered is %d\n", anint);
printf("\nThe char entered is %c\n", achar);

It allows entering an integer, then skips the second scanf completely, this is really strange, as when I swap the two (the char scanf first), it works fine. What on earth could be wrong?

Answer 1

It doesn't skip the second scanf(); the second scanf() reads the newline left behind by the first scanf(). Most format codes skip white space; the %c format does not skip white space.

Answer 2

When reading input using scanf, the input is read after the return key is pressed but the newline generated by the return key is not consumed by scanf, which means the next time you read a char from standard input there will be a newline ready to be read.

One way to avoid is to use fgets to read the input as a string and then extract what you want using sscanf as:

char line[MAX];

printf("\nEnter any integer:");
if( fgets(line,MAX,stdin) && sscanf(line,"%d", &anint)!=1 ) 
   anint=0;

printf("\nEnter any character:");
if( fgets(line,MAX,stdin) && sscanf(line,"%c", &achar)!=1 ) 
   achar=0;

Another way to consume the newline would be to scanf("%c%*c",&anint);. The %*c will read the newline from the buffer and discard it.

You might want to read this:

C FAQ : Why does everyone say not to use scanf?

Answer 3

The other answers are correct - %c does not skip whitespace. The easiest way to make it do so is to place whitespace before the %c:

scanf(" %c", &achar);

(Any whitespace in the format string will make scanf consume all consecutive whitespace).

Answer 4

Try also _flushall() after each printf call. . Basically, by default MS’s C++ buffers stream output, and the the flushing causes the output stream to empty.