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Message: [QUOTE="nrk, post: 1696052"] Ah!! The infamous '\n' in the scanf format specifier. The problem with this is that it consumes *any* amount of white space (space, tabs and newlines), not just a single newline. So what happens is that after you hit say 10\n, the '\n' in your format specifier forces scanf to keep waiting for more input (since it consumes any amount of whitespace, it needs to wait till it finds a non-whitespace character in the input stream). The only way to make the scanf return is by giving a non-whitespace input after your normal input. However, note that it won't consume the non-whitespace input that you gave but leave it behind in the input stream, so that your next scanf (the one with "%1.0\n") can consume that. Of course, the same problem is again posed by the '\n' in your other format specifier. What is the fix? Unfortunately, no straightforward one exists. The best solution is to split up your scanf's. One to read the integer/float and any remaining characters (discarded) in the input line, and the next to consume the newline in the input stream. For instance: int ret = scanf("%d%*[^\n]", &total); if ( ret < 1 ) { /* ret == 0 means invalid input */ /* ret == EOF means error reading input stream or end of input */ /* handle appropriately */ } if ( !feof(stdin) ) getchar(); /* discard newline */ A fix along similar lines can be applied to your other scanf as well. Variable length arrays are a C99 feature. To be portable to C89 compilers, you will need to use dynamic memory allocation here instead of using a VLA. HTH, -nrk. PS: Rest assured, your original code only reads the first <total> floats from the input stream. You can easily construct an input to check this out (make the least number the last one you enter, and watch as the minimum printed by the program doesn't match this value). [i][i][i][/i][/i][/i] [/QUOTE]

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