fgets falls through without taking input

Question

whenever I call this function it fails to give the use a chance to enter the string. Not sure why pretty sure my syntax is spot on. think I has something to do with the newline but I don't know how to get rid of it

 void serchName(dealers_t *ptr, int numDealers) { char dealerName[NAME_LEN]; int index; printf("please enter the dealer's name:"); fgets(dealerName, sizeof(dealerName), stdin); system("PAUSE"); for (index = 0; index < numDealers; index++, ptr++) { if (strcmp(dealerName, ptr->name) == 0) { printf("MATCH FOUND:%s\n%s\n%s\n%i\n%s\n", ptr->name,ptr->city,ptr->state,ptr->zip,ptr->phone); } } }

You have something in stdin from before the call. Make sure you don't. — this
– this, Commented Apr 30, 2014 at 20:15
Fgets also reads the terminal '\n' into the buffer, so the strcmp() will fail (unless the ptr->name also has an '\n' at its end) — wildplasser
– wildplasser, Commented Apr 30, 2014 at 21:22

chux · Accepted Answer · 2014-04-30 22:24:42Z

You certainly have some '\n' left-over from previous I/O activity.

It is best to use fgets() and not mixed with scanf() in the same program.

But since I can not see the other code, suggest the following to consume a left-over '\n'.

printf("please enter the dealer's name:"); int ch; ch = fgetc(stdin); if (ch != '\n') ungetc(ch, stdin); if (fgets(dealerName, sizeof(dealerName), stdin) == NULL) Handle_EOForIOError();

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