int a=5 double d=12345.6789 printf("%*.*lf\n",a+5,a-3,d) 请问答
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:综合作业 时间:2024/07/05 22:39:49
int a=5 double d=12345.6789 printf("%*.*lf\n",a+5,a-3,d) 请问答案是什么?
![int a=5 double d=12345.6789 printf(](/uploads/image/z/18760889-65-9.jpg?t=int+a%3D5+double+d%3D12345.6789+printf%28%22%25%2A.%2Alf%5Cn%22%2Ca%2B5%2Ca-3%2Cd%29+%E8%AF%B7%E9%97%AE%E7%AD%94)
%*.*表示最小宽度和精度由参数提供,所以最小宽度是a+5也就是10,精度是a-3也就是保留两位小数且要四舍五入.
C语言 int a;int b; }d[3] ={{1,4},{2,5},{6,7}}; printf("%d\n",d
{ int a[]={1,2,3,4,5,6}; int*p; p=a; printf("%d\n",*p); prin
int a=10;f1(){int a=20;printf("%d",a);}f2(){printf("%d",a);}
main() {int a[10]={1,2,3,4,5}; printf("%d\n",a[6]);}输出的结果是啥,
#include main() { int a=5,b=4,c=3,d; d=(a>b>c); printf("%d\n
Int a=1; Int *p; p=&a; printf(“%d\n”,*p); 和Int a=1; Int*p; *
以下程序的输出结果是 int a=5,b=4,c=6,d; printf("%d\n",d=a>b?(a>c?a:c)b
int a=5;printf ("%%d",a);输出的为什么是%d呀具体点,
main() { int a=5,b=4,x,y; x=a++*a++*a++; printf(“a=%d,x=%d\n
void main( ) { int a=5,b=8; printf(“a=%d b=%d\n”,a,b) ; a=a+
main( ) {unsigned int a=3,b=10; printf("%d/n",a1); }
#include main() {unsigned int a=65535; printf("a=%d\n",a); }