你可能感兴趣的试题
(p++)->num p++ (*.num P=&stag
double x[5] =2.0,4.0,6.0,8.0,10.0; int y[5] =0,1,3,5,7,9; char c 1[ ] ='1','2','3','4','5'; char c 2 [C] =,'/x10','/xa','/x8';
double x[5]=2.0,4.0,6.0,8.0,10.0; int y[5]=0,1,3,5,7,9; char cl[]='1','2','3','4','5'; char c2[]='/x10','/xa','/x8';
double x [5]={ 2.0,4.0,6.0,8.0,10.0 }; int y [5]={ 0,1,3,5,7,9 }; char cl[ ]={'1','2','3','4','5' }; char c2[ ]:{ '/xlO','/xa','/x8' };
(p++)->num p++ (*p).num p=&stuage
double x[5]=1.0,2.0,3.0,4.0,5.0; int y[5]=0,1,2,3,4,5; char c1[]='1','2','3','4','5'; char c2[]='a','b','c';
double x[5]={2.0,4.0,6.0,8.0,10.0}; int y[5]={0,1,3,5,7,9}; char c1[]={'1','2','3','4','5'}; char c2[]={'/x10','/xa','/x8'};
在不同函数中可以使用相同名字的变量 形式参数是局部变量 在函数内定义的变量只在本函数范围内在效 在函数内的复合语句中定义的变量在本函数范围内在效
ps=(char*)malloc(8); ps=(char *)malloc(sizeof(char)* 8); ps=(char*)calloc(8,sizeof(char)) ps=8*(char*)malloc(sizeof(char))
double x[5] =2.0,4.0,6.0,8.0,10.0; int y[5] =0,1,3,5,7,9; char c 1[ ] ='1','2','3','4','5'; char c 2 [C] =,'/x10','/xa','&
doublex[5]={2.0,4.0,6.0,8.0,10.0}; inty[5]={0,1,3,5,7,9}; charcl[]={'1','2','3','4','5'}; charc2[]:{'/x1O','/xa','/x8'};
double x[5]=2.0,4.0,6.0,8.0,10.0; int y[5.3]=0,1,3,5,7,9; charc/[]='1','2','3','4','5'; char c2[]='/x10','/xa','/x8';
(p++) ->num p++ (*p) .num p=&stu.age
double x[5]=2.0,4.0,6.0,8.0,10.0; int y[5.3]=0,1,3,5,7,9; char c1[]='1','2','3','4','5'; char c2[]='/x10','/xa','/x8';
double x[5] ={2.0,4.0,6.0,8.0,10.0}; int y[5] ={0,1,3,5,7,9}; char c 1[ ] ={'1','2','3','4','5'}; char c 2 ={,'/x10','/xa','/x8'};
ps=(char*)malloc(8); ps=(char *)malloc(sizeof(cha* 8); ps=(char*)calloc(8,sizeof(cha) ps=8*(char*)malloc(sizeof(cha)