![下列程序段的输出结果是 【31】 。 char *s[]={"East","West","South","North"}; char **p=s; printf("?,第1张 下列程序段的输出结果是 【31】 。 char *s[]={"East","West","South","North"}; char **p=s; printf("?,第1张](/aiimages/%E4%B8%8B%E5%88%97%E7%A8%8B%E5%BA%8F%E6%AE%B5%E7%9A%84%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C%E6%98%AF+%E3%80%9031%E3%80%91+%E3%80%82+char+%2As%5B%5D%3D%7B%26quot%3BEast%26quot%3B%2C%26quot%3BWest%26quot%3B%2C%26quot%3BSouth%26quot%3B%2C%26quot%3BNorth%26quot%3B%7D%3B+char+%2A%2Ap%3Ds%3B+printf%28%26quot%3B%EF%BC%9F.png)
选择答案C。
for(i=0i<4i++)s+=aa[i][1]
这个循环之后,s就是数组aa的一到四排第二个元素之和,
即:s=aa[0][1]+aa[1][1]+aa[2][1]+aa[3][1]=2+6+9+2=19
故输出结果为:19
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![下列程序段的输出结果是 【31】 。 char *s[]={"East","West","South","North"}; char **p=s; printf("?,第1张 下列程序段的输出结果是 【31】 。 char *s[]={"East","West","South","North"}; char **p=s; printf("?,第1张](/aiimages/%E4%B8%8B%E5%88%97%E7%A8%8B%E5%BA%8F%E6%AE%B5%E7%9A%84%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C%E6%98%AF+%E3%80%9031%E3%80%91+%E3%80%82+char+%2As%5B%5D%3D%7B%26quot%3BEast%26quot%3B%2C%26quot%3BWest%26quot%3B%2C%26quot%3BSouth%26quot%3B%2C%26quot%3BNorth%26quot%3B%7D%3B+char+%2A%2Ap%3Ds%3B+printf%28%26quot%3B%EF%BC%9F.png)
选择答案C。
for(i=0i<4i++)s+=aa[i][1]
这个循环之后,s就是数组aa的一到四排第二个元素之和,
即:s=aa[0][1]+aa[1][1]+aa[2][1]+aa[3][1]=2+6+9+2=19
故输出结果为:19
欢迎分享,转载请注明来源:内存溢出
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