c - 格式字符串未使用数据参数 - 输出长整型

Question

好吧，这是一个 C 程序...即使我已经正确放置了它，我还是遇到了这个问题。长毫秒，在打印中我有“l”。现在，x 是一个全局 int 变量，用户通过命令行参数输入 x，然后 atoi() 作为参数。

此问题发生在函数中。我想知道我是否输出错误。在尝试输出之前将 int x 设置为这个 long ms 变量时，我是否应该进行类型转换？我很困惑并试图输出毫秒。

    struct timespec now;
 clock_gettime(CLOCK_REALTIME, &now);
 long ms;
 x = ms;
 ms = round(now.tv_nsec / 1.0e6);

 fprintf(stdout, "l: flip\n", ms);

Answer 1

我已经用我所看到的内容注释了您的代码。

struct timespec now;
clock_gettime(CLOCK_REALTIME, &now);
long ms;
x = ms;  
// <== this is assigning a variable 'ms' to the variable 'x'
// however, the variable 'ms' is not initialized 
// this will cause the compiler to raise a warning message
// about using an uninitialized variable
// this is (for the above reason) undefined behavior

ms = round(now.tv_nsec / 1.0e6); 
// <-- initialized the variable 'ms'
// but it is too late for the prior statement
// AND the compiler will not be happy about the 'implied' 
// conversion from the returned type from 'round()' 
// which is 'double' to the type 'long int' of the variable 'ms'
// The compiler will be happy if the line is written like this:
ms = (long int)round( now.tv_nsec / 1.0e6 );


fprintf(stdout, "l: flip\n", ms); 
// <-- will cause the compiler to complain, as you already know

如果代码是:

fprintf(stdout, "%l: flip\n", ms); 
// <-- will also cause the compiler to complain about a invalid format specifier

如果代码使用与变量“ms”(long int)的“类型”匹配的有效格式说明符

fprintf(stdout, "%ld: flip\n", ms); 
// <-- now compiler happy

格式说明符“%ld”表示“长整数”

fprintf()(或 printf())的手册页告诉您所有可用的有效格式说明符及其用途。