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Why doesnt the call scanf work?

Q

Why doesn't the call scanf("%d", i) work?

✍: Guest

A

The arguments you pass to scanf must always be pointers: for each value converted, scanf ``returns'' it by filling in one of the locations you've passed pointers to. To fix the fragment above, change it to scanf("%d", &i) .

2015-11-02, 1781👍, 0💬

.NET C C++ CSS DBA General HTML Java JavaScript JSP JUnit MySQL Networking Oracle Perl Perl PHP PL/SQL RSS Software QA SQL Server Struts Unix Windows XHTML XML