Why is my program not outputting anything with printf?

Question

I am a student in a network programming class and I am writing a UDP client that connects to three servers to do a file transfer. So far, Ive gotten the client to communicate with the first 2 servers but this 3rd copy of the server is not working. Its not even printing out the printf debug statements I put in. What is going on? Is there something wrong with Unix? The first two servers are exactly the same as this one with changes in port/ip address and they work perfectly fine.

#include <stdlib.h> #include <arpa/inet.h> #include <netinet/in.h> #include <stdio.h> #include <errno.h> #include <sys/socket.h> #include <sys/types.h> #include <strings.h> int main(int argc, char**argv) { printf("i will destroy you"); int o = 1; int sockfd,n; struct sockaddr_in servaddr,cliaddr; socklen_t len; char mesg[1000]; printf("im confused"); sockfd=socket(AF_INET,SOCK_DGRAM,0); bzero(&servaddr,sizeof(servaddr)); servaddr.sin_family = AF_INET; servaddr.sin_addr.s_addr=inet_addr("127.0.0.1"); servaddr.sin_port=htons(3454); bind(sockfd,(struct sockaddr *)&servaddr,sizeof(servaddr)); printf("yo wtf"); if (setsockopt(sockfd, SOL_SOCKET, SO_REUSEADDR, &o, sizeof(o)) < 0){ perror("setsockopt failed"); exit(0); } printf("here1"); while(1) { len = sizeof(cliaddr); printf("here"); n = recvfrom(sockfd,mesg,1000,0,(struct sockaddr *)&cliaddr,&len); sendto(sockfd,mesg,n,0,(struct sockaddr *)&cliaddr,sizeof(cliaddr)); printf("-------------------------------------------------------\n"); mesg[n] = 0; printf("Received the following:\n"); printf("%s",mesg); printf("-------------------------------------------------------\n"); } }

For a start, put \n at the end of your printf strings. You may not be seeing them because they're buffered. — paxdiablo
– paxdiablo, Commented Feb 20, 2013 at 3:45
If you want output messages to appear, make sure they end with a newline. Your first two printf() statements do not end with a newline; their output won't appear until after you output a newline, or use fflush(stdout) (or print so much data on a single line that it no longer fits in the buffer in the FILE * for stdout). — Jonathan Leffler
– Jonathan Leffler, Commented Feb 20, 2013 at 3:45
printf(3) output normally is buffered, it is only written out when the buffer (largeish) fills. You can force output with fflush(stdout). — vonbrand
– vonbrand, Commented Feb 20, 2013 at 3:46

Joseph Quinsey · Accepted Answer · 2013-02-20 03:45:32Z

1

You probably need to change e.g. printf("i will destroy you") to printf("i will destroy you\n"). Or see setbuf(). Or use fflush(stdout).

answered Feb 20, 2013 at 3:45

Joseph Quinsey

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6 Comments

James Le Over a year ago

it does print now. why is the newline character needed? sometimes in my other programs i leave it out but it still shows up somewhere before the prompt in the terminal.

Joseph Quinsey Over a year ago

scanf, etc, which uses stdin, is tied to stdout, so that prompts don't get garbled.

Joseph Quinsey Over a year ago

One minor point: the line mesg[n] = 0; is not quite right. n could possibly be 1000, and so the nul is put past the end of the array. (Or, of course, n could be -1, if you wished to check for errors.)

Joseph Quinsey Over a year ago

See stackoverflow.com/q/2123528/318716, Does reading from stdin flush stdout? But the answers there seem to disagree with my comment.

phoeagon Over a year ago

does a \n always guarantee a flush?

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Doug Currie · Accepted Answer · 2013-02-20 03:45:24Z

0

You are not flushing stdout.

Alternatively, use stderr as in

fprintf(stderr,"My message\n");

answered Feb 20, 2013 at 3:45

Doug Currie

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