#include main( ) {int x,y; scanf("%2d%ld",&x,&y); printf("%d\n",x+y); } 输入1234567程序运行结果是多少?为什么长整形的y就是34567呢?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 20:47:10
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#include main( ) {int x,y; scanf("%2d%ld",&x,&y); printf("%d\n",x+y); } 输入1234567程序运行结果是多少?为什么长整形的y就是34567呢?
#include main( ) {int x,y; scanf("%2d%ld",&x,&y); printf("%d\n",x+y); } 输入1234567
程序运行结果是多少?
为什么长整形的y就是34567呢?
#include main( ) {int x,y; scanf("%2d%ld",&x,&y); printf("%d\n",x+y); } 输入1234567程序运行结果是多少?为什么长整形的y就是34567呢?
scanf("%2d%ld",&x,&y);
当输入1234567时,%2d,取前2位,放入x,即x=12;%ld【这个是long的l,不要看成1,如果是1,只能取一位,3】,取后面的34567,放入y,即y=34567
printf("%d\n",x+y);,输出34567+12=34579