int a=1234;
    printf("a=%d\n",a);    //a=1234
    printf("a=%2d\n",a);   //a=1234    超过2位，按实际输出
    printf("a=%6d\n",a);   //a=  1234  不足6位，右对齐
    printf("a=%06d\n",a);  //a=001234  不足6位，前面补0
    printf("a=%-6d\n",a);  //a=1234    '-'左对齐
    

int* i=&a;
    printf("i=%p\n",i);    //i=0012FF44  输出指针的值，即地址

float m=8888.8888;     //float 单精度型浮点数 有效位数是6位或7位,根据不同的浮点数会有不同
float m1=8888.8888f;    //在后面加上f或F，编译警告：truncation from'const double'to 'float'
//编译器默认浮点数为double
float m2=8888.888f;  
double n=8888.8888;
double n1=8888888888.88888888;  //double 双精度型浮点数 有效位数是15位
    printf("m=%f\n m1=%f\n m2=%f\n n=%lf\n n1=%f\n",m,m1,m2,n,n1); // m=8888.888672
//  m1=8888.888672     
//  m2=8888.887695
// n=8888.888800
// n1=8888888888.888889
//%f的默认输出小数位数就是6位不管有没有l
/*printf的%f说明符的确既可以输出float型又可以输出 double型。 根据"默认参数提升"规则（在printf这样的函数的
    可变参数列表中 ，不论作用域内有没有原型，都适用这一规则）float型会被提升为double型。因此printf()只会看到
    双精度数。严格地讲，%lf在printf下是未定义的，但是很多系统可能会接受它。要确保可移植性，就要坚持使用%f。*/

    printf("m4=%4.2f\n",m);   //宽度总共4位，小数两位，小数点一位，整数一位，这里整数超过宽度规定，按实际整数位输出
    printf("m5=%9.6f\n",m);   //浮点数小数部分不足6位，右对齐
    printf("m6=%9.2f\n",m);   //整数部分不足6位，右对齐；小数部分超过2位，四舍五入

char c[20]="Hello,world!";
    printf("c=%s\n",c);    
    printf("c=%6.9s\n",c);  //c=Hello,wor  6.9s表示输出一个长度长度不小于6且不大于9的字符串。若大于9, 则第9个字符以后的内容将被删除。
----------------
What should a compiler on a 64bit platform do with the following:


unit64_t    x=1234567;

printf("%llu", x);


On 32Bit unit64_t is usually a macro for "unsigned long long" so above

example is perfectly right.

On 64Bit unit64_t is usually a macro for "unsigned long" so above code might

cause a warning.


Of course you can use PRIu64 to avoid that, but many docs on printf tell

you:


"The optional character sequence ll, specifying that a following d, i, o, u,

x, or X conversion corresponds to a **quad** int or unsigned **quad** int

argument, or that a following n conversion corresponds to a pointer to a

quad int argument.  The use of q has been deprecated as conversion

character."


A quad, however, is 64bit on every platform and hence %llu should print

uint64_t without a warning.


What do you think? Is this right or wrong?

What is the real meaning of %llu:

 - print a "long long" variable?

 - or print a 64bit varibale?