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| (2 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
== 1-1 == | == 1-1 == | ||
int main() { | int main() { | ||
int number | int number[10] = { 1,3,5,7,9,11,13,15,17,19 }; | ||
int i, * p; | int i, * p; | ||
| Line 13: | Line 13: | ||
int main() { | int main() { | ||
int number | int number[10] = { 1,3,5,7,9,11,13,15,17,19 }; | ||
int i, * p; | int i, * p; | ||
| Line 19: | Line 19: | ||
for (i = 0; i < 10; i++) { | for (i = 0; i < 10; i++) { | ||
printf("%d\n", p | printf("%d\n", p[i]); | ||
} | } | ||
return 0; | return 0; | ||
| Line 26: | Line 26: | ||
int main() { | int main() { | ||
int number | int number[10] = { 1,3,5,7,9,11,13,15,17,19 }; | ||
int i; | int i; | ||
int *p = &number | int *p = &number[0]; | ||
for (i = 0; i < 10; i++) { | for (i = 0; i < 10; i++) { | ||
| Line 36: | Line 36: | ||
} | } | ||
== 2-1 == | == 2-1 == | ||
int main() { | |||
int number[3][4] = { {1,3,5,7},{9,11,13,15},{17,19,21,23} }; | |||
int i, j,*p; | |||
for (i = 0; i < 3; i++) { | |||
for (j = 0; j < 4; j++) { | |||
printf("%d\n", number[i][j]); | |||
} | |||
} | |||
return 0; | |||
} | |||
== 2-2 == | == 2-2 == | ||
int main() { | |||
int number[3][4] = { {1,3,5,7},{9,11,13,15},{17,19,21,23} }; | |||
int i, j,*p=&number[0][0]; | |||
for (i = 0; i < 3; i++) { | |||
for (j = 0; j < 4; j++) { | |||
printf("%d\n", *p++); | |||
} | |||
} | |||
return 0; | |||
} | |||
Latest revision as of 01:08, 27 March 2026
1-1
int main() {
int number[10] = { 1,3,5,7,9,11,13,15,17,19 };
int i, * p;
for (i = 0; i < 10; i++) {
printf("%d\n", *(number + i));
}
return 0;
}
1-2
int main() {
int number[10] = { 1,3,5,7,9,11,13,15,17,19 };
int i, * p;
p = number;
for (i = 0; i < 10; i++) {
printf("%d\n", p[i]);
}
return 0;
}
1-3
int main() {
int number[10] = { 1,3,5,7,9,11,13,15,17,19 };
int i;
int *p = &number[0];
for (i = 0; i < 10; i++) {
printf("%d\n", *p++);
}
return 0;
}
2-1
int main() {
int number[3][4] = { {1,3,5,7},{9,11,13,15},{17,19,21,23} };
int i, j,*p;
for (i = 0; i < 3; i++) {
for (j = 0; j < 4; j++) {
printf("%d\n", number[i][j]);
}
}
return 0;
}
2-2
int main() {
int number[3][4] = { {1,3,5,7},{9,11,13,15},{17,19,21,23} };
int i, j,*p=&number[0][0];
for (i = 0; i < 3; i++) {
for (j = 0; j < 4; j++) {
printf("%d\n", *p++);
}
}
return 0;
}