Chopper small scale (two)

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1. nested loop, printing characters in the following format

#
##
###
####
#####

嵌套循环解释：
for(){
	for(){
	}
}

Solution: (* Note prov-for.c)

for(i=0;i<5;i++){
	for(j=0;j<=i;j++){
		printf("#");
	}
	printf("\n");
}

2. Use switch calculating a score, the score input, the output level

90-100	A
70-89	B
60-69	C
<60	D

Solution: (* prov-switch.c)

int i;
printf("input sorce:\n");
scanf("%d",&i);
i=i/10;
switch(i){
	case 10:
	case 9:
		printf("A\n");
		break;	
	case 8:
	case 7:
		printf("B\n");
		break;
	case 6:
		printf("C\n");
		break;
	default:
		printf("D\n");
		break;

}

3. Design a function chline (ch, i, j), i-th row specified character printing, j column.

Example: chline ( 'a', 3 , 2) Output:
AA
AA
AA

Solution: (* Note prov-chline.c)

viod chline(char ch,int i,int j){
	int a;
	int b;
	for(a=0;a<i;a++){
		for(b=0;b<j;b++){
			printf("%s",ch);
		}
		printf("\n");
	}
}

4. Write a function mymax (int array []), returns the maximum value stored in the array of type int.

Solution: (* prov-max.c)

int mymax(int aray[],int n){
	int i;
	int max;
	max = array[0];
	for(i=0;i<n;i++){
		if(max<array[i]){
			max=array[i];
		}
	}
	return max;
}

5. Design a GetMemory function, the application 100 bytes of memory, and the memory is transmitted back

Solution: (** Note # prov-getmemory.c)

method one:

char *getmemory(){
	char *p=malloc(100);
	return p;
}

Method two :( achieved by passing the two pointers)

void getmemory(char **p){
	*p=malloc(100);
}

summary: Compare ignorant of the fifth question, the understanding of a pointer and secondary pointer is not accurate enough place, resulting in using up more confusion; All in all, also to be strengthened.