题目内容

当运行以下程序时输入OPEN THE DOOR,则输出结果是 #include #include char fun(char *c) { if(*c <='Z' && *c>= ‘A’ ) *c -= ‘A’ - ‘a’; return *c; } main() { char s[80], *p = s; gets(s); while(*p) { *p = fun( p ); putchar(*p); p++; } putchar(‘\n’); }

A. oPEN tHE dOOR
B. Open The Door
C. open the door
D. OPEN THE DOOR

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以下程序的输出结果是 #include #include void fun( char *w, int m) { char s, *p1, *p2; p1 = w; p2 = w+m-1; while(p1 < p2) { s = *p1 ++; *p1 = *p2--; *p2 = s; } } main() { char a[] = “ABCDEFG”; fun(a, strlen(a)); puts(a); }

AGADAGA
B. AGAAGAG
C. GFEDCBA
D. GAGGAGA

以下程序的输出结果是 main() { char ch[2][5] = {“6937”, “8254”}, *p[2]; int i, j, s = 0; for(i = 0; i < 2; i++) p[i] = ch[i]; for(i = 0; i < 2; i++) for(j = 0; p[i][j] > ‘\0’ && p[i][j] <= ‘9’; j += 2) s = 10 * s+ p[i][j] - ‘0’; printf(“%d\n”, s); }

A. 69825
B. 693825
C. 6385
D. 63825

以下程序的输出结果是 sub(int *s, int *y) { static int t = 3; *y = s[t]; t--; }main() { int a[] = {1, 2, 3, 4}, i, x = 0; for(i = 0; i < 4; i++) { sub(a, &x); printf(“%d ”, x); } printf(“\n”); }

A. 4 4 4 4
B. 0 0 0 0
C. 1 2 3 4
D. 4 3 2 1

以下程序的输出结果是 main() { int k = 2, m = 4, n = 6; int *pk = &k, *pm = &m, *p; *(p = &n) = *pk * (*pm); printf(“%d\n”, n); }

A. 8
B. 10
C. 6
D. 4

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