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중간고사 feedback. 2011. 10. 25 최 윤 정. Point. Binary / Hexadecimal / Decimal Grammar in C language - %d %f %lf %u %x %p … - *, & Floating Point Format Stack Structure. Page 3 : Chap.2 & C 의 기본문법 ( 1/4). 1) 0x502c + 64 = 0x502c + 0x40 = 0x506c, %x 출력형태는 506c

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  1. 중간고사 feedback 2011. 10. 25최윤 정

  2. Point • Binary / Hexadecimal / Decimal • Grammar in C language - %d %f %lf %u %x %p …- *, & • Floating Point Format • Stack Structure

  3. Page 3 : Chap.2 & C의 기본문법(1/4) 1) 0x502c + 64 = 0x502c + 0x40 = 0x506c, %x 출력형태는 506c 2) 0x50da + 0x502c = 0xa106 , %x 출력형태는 a106 3) x = 0xFFFFFF0, x<<3 ? • X = 00001111……11110000 , X<<3 = 00001111……….111 1000 0000, ∴ 7FFFFFF80 4) unsigned int x = 0xFFFFFFF0, x>>2 ? • Arithmetic shift vs. Logical shift • X = 1111 ………. 0000, X >> 2 = 001111……….1111 0000,∴3FFFFFFC

  4. Page 3 : Chap.2 & C의 기본문법(2/4) 5) int x = 0x15213F10 >> 4; //즉 0x015213F1 char y = (char) x // y = 0xF1 unsigned char z = (unsigned char) x; z = 0xF1 • F1 = 11110001, • signed = -x7(27) + (x6(26) + … x0(20)) = -128 +(64+32+16+1)= -15 • unsigned = x7(27) + x6(26) + … x0(20) = 128+64+32+16+1 = 241 6) float x = 31/8 ; float y = 31/8.0; • Arithmetic : 31/8 = 3.875, but, integer31 / interger8 = 3, (∵ 11111 >> 3) • (float) 31/ 8.0 = 3.875

  5. Page 3 : Chap.2 & C의 기본문법(3/4) 7) n-bit signed integer의 범위 : -2n-1 ~ 0 ~ 2n-1 - 1 n-bit unsigned integer의 범위 : 0 ~ 2n – 1 • Two’s complement에서, 즉, signed integer로 -1과 1을 표현하는 필요한 bit size • -1 : -21-0 n = 1, 1bit • 1 : 2-1 = 22-1 - 1 = 1, 2bit 8) 4 page에서 9) 15213U > -1 • 15213U >(unsigned) -1 ,15213U > FFFFFFFF ? false ∴ 0. 10) INT_MIN == TMIN • 1000 …. 0000, 32bit에서, 800000000 • -INT_MIN = INT_MAX + 1  INT_MIN 즉, 800000000

  6. Page 3 : Chap.2 & C의 기본문법(4/4) 11) , 12) : a • - n = ~ n + 1, • ~(0000101101) + 1 = 1111010010 +1 = 1111010011 13) %eax 14) sizeof(char) + sizeof(int) + sizeof(void *) + sizeof(int *) • IA32 : 1 + 4 + 4 + 4 = 13, • Intel x86_64 : 1 + 4 + 8 + 8 = 21 , sizeof(int) =4, sizeof(long int) = 8. • 그러나, linux의 버전과 compiler별 차이를 분명히명시하지 않았던 점으로 인하여 sizeof(int) = 8 로 계산한 1 + 8 + 8 + 8 = 25 의 답도 인정함. 15) Lea vs. Mov, • Lea: 0x10(%eax, %ecx, 4) , %eax %eax = 16+ %eax + 4*%ecx • Mov: 0x10(%eax, %ecx, 4) , %eax %eax = *(16 + %eax + 4*%ecx) • 16) • #include<stdio.h> • int main(){ • int x = 0; • printf("Please input an integer:"); • scanf("%d", x ); • printf("%d", (!!x)<<31); • }

  7. Page 4 : Chap.2 – Floating Point Format • 7-bit floating point representation based on the IEEE floating point • There is a sign bit in the most significant bit. • The next 3 bits are the exponent. The exponent bias is 3. • The last 3 bits are the fraction. V = (-1)s * M * 2E M is the significand and E the integer value of the exponent 0 000 000 • E = exp - Bias (norm.) • E = 1 - Bias ( denorm.) exp frac. 2 와 3/8 = 10.011 (-1) 0 * 1.0011 * 21 따라서, s = 0 E = 1 M = 1.0011, exp = 3 +1 = 4 frac. = 010 ( ∵ 소수부0011의 3bit를취할때,round-to-even 적용)

  8. Page 3. 8) ¼ - bitrepresentation in IEEE 754 • IEEEE 754 = 1 sign bit, 8 exp bits , 23 frac. bits • Bias = 28-1 – 1 = 27– 1 = 127 , sign bit = 0. • ¼ = 0.01 • (-1)0 * 1.0 * 2 -2 • exp = Bias + E = 127 + (-2) = 125, 0111 1101 • frac = 0000 …. 0000 • ∴0 01111101 0000…0000 • = 3E8 0 0 0 0 0

  9. Stack “Bottom” Increasing Addresses Stack Grows Down Stack Pointer: %esp Stack “Top” IA32 Stack • Region of memory managed with stack discipline • Grows toward lower addresses • Register %esp contains lowest stack address • address of “top” element • Push : %esp = %esp – 4 • Pop : %esp = %esp+ 4 • Call Procedure : • pushnext address(Return address) . . . . 0x110 • ex) call 0x8048b90 • (Next addr = 0x8048553) 0x10c 0x104 0x108 123 %esp 0x108 0x8048b90 0x104 0x8048553 %eip 0x804854e

  10. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x108 . . . . 0x110 %ebp 0x804854e 0x10c %ebx i 0x108 %esp %edx j %eax 0x804b1204 %eip 0x804b1100

  11. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x108 . . . . 0x110 %ebp 0x804854e j 0x10c %ebx i 0x108 i %esp %edx j %eax 0x804b1204 %eip 0x804b1200

  12. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x104 . . . . 0x110 %ebp 0x804854e j 0x10c %ebx i 0x108 i 0x104 0x804b1204 %esp %edx j %eax 0x804b1204 %eip 0x804b90

  13. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x100 . . . . 0x110 %ebp 0x804854e j 0x10c %ebx i 0x108 i 0x104 0x804b1204 %edx j 0x804854e %esp 0x100 %eax ... //setupcode in func. print pushl %ebp movl %esp, %ebp … %eip 0x804b90+a

  14. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x100 . . . . 0x110 %ebp 0x100 j 0x10c %ebx i 0x108 i 0x104 0x804b1204 %edx j %ebp 0x804854e 0x100 %esp %eax ... //setupcode in func. print pushl %ebp movl %esp, %ebp … pop %ebp %eip 0x804b90+b

  15. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x104 . . . . 0x110 %ebp 0x804854e j 0x10c %ebx i 0x108 i 0x104 0x804b1204 %esp %edx j %eax ... //return from func. print … pop %ebp %eip 0x804b90+c

  16. 9*9단 inline- assembly code 에서 void print(int x, int y); .. ".Lp3 : " "movl %1, 4(%%esp) " "movl %0, (%%esp) " "call print"  //0x804b90 "movl 4(%%esp), %1" "movl (%%esp), %0 " ... :"r" (i), "r" (j).. call 8048b90 %esp 0x108 . . . . 0x110 %ebp 0x804854e j 0x10c %ebx i 0x108 i %esp %edx j %eax ... %eip 0x804b1204

  17. voidgugu_no_loop(){ • int i=0, j=0; • __asm__ __volatile__( • "movl $0, %0 \n\t" • "movl $0, %1 \n\t" • ".Lp1 : ""cmpl $9, %0 \n\t" • "jge .End \n\t" • "incl %0 \n\t" • ".Lp2 : ""cmpl $9, %1 \n\t" • "jl .Lp4 \n\t" • "movl $0, %1 \n\t" • "movl %1, 4(%%esp) \n\t" • "movl %0, (%%esp) \n\t" • "call println \n\t" • "movl 4(%%esp), %1 \n\t" • "movl (%%esp), %0 \n\t" • "jmp .Lp1 \n\t" • ".Lp3 : " "movl %1, 4(%%esp) \n\t" • "movl %0, (%%esp) \n\t" • "call print \n\t" • "movl 4(%%esp), %1 \n\t" • "movl (%%esp), %0 \n\t" • "jmp .Lp2 \n\t" • ".Lp4 : ""incl %1 \n\t" • "jmp .Lp3 \n\t" • ".End : ""movl $0, %1 \n\t" • : • :"r" (i), "r" (j) • ); • } #include<stdio.h> voidgugu_for(); voidgugu_no_loop(); voidprintln(){ printf("\n"); } void print(int x, int y){ printf("%2d x %2d = %2d \n", x, y, x*y); } voidgugu_for(){ inta, sum, i, j; a = 1; for( i=1; i<=9; i++){ for(j=1; j<=9; j++){ __asm__ __volatile__( "movl %1, %0 \n\t" "imull %2, %0 \n\t" "movl %2, 4(%%esp) \n\t" "movl %1, (%%esp) \n\t" "call print" : "=g" (sum) :"r" (i), "r" (j) ); // printf("%d x %d = sum=%d \n", i, j, sum); } printf("\n"); } }

  18. %ebp %ebp %ebp %ebp %esp %esp %esp %esp Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x300 Stack 1 • pushl %ebp • movl %esp,%ebp • movl 8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret … 0x10c 2 0x200 %esp 0x108 0x100 func1 0x104 %ebp 0x100 func2 func3 0x100 %eax 300 200 %edx Rtnaddr Old ebp %ebp %esp

  19. Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x300 1 • pushl %ebp • movl %esp,%ebp • movl 8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 0x10c 2 0x200 %esp 0x108 0x100 0x104 %ebp 0x100 0x100 %eax *(0x100+c) 300 300 200 %edx *(0x100+8) 200 Rtnaddr Old ebp %ebp %esp

  20. 0x300 Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x200 • pushl %ebp • movl %esp,%ebp • movl8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 1 0x10c 2 %esp 0x108 0x100 0x104 %ebp 0x100 0x100 %eax *(0x100+c) 300 300 200 %edx *(0x100+8) 200 Rtnaddr Old ebp %ebp %esp

  21. 0x300 Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x200 • pushl %ebp • movl %esp,%ebp • movl8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 1 0x10c 2 %esp 0x108 0x100 0x104 %ebp 0x100 0x100 %eax *(0x100+c) 300 300 200 %edx 2 Rtnaddr Old ebp %ebp %esp

  22. 0x300 Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x200 • pushl %ebp • movl %esp,%ebp • movl8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 1+2 0x10c 2 %esp 0x108 0x100 0x104 %ebp 0x100 0x100 %eax *(0x100+c) 300 300 200 %edx 2 Rtnaddr Old ebp %ebp %esp

  23. 0x300 Page 4 : Chap. 3 – Stack&Assembly코드의 이해 0x200 • pushl %ebp • movl %esp,%ebp • movl8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 1+2 0x10c 2 %esp 0x108 0x100 0x104 %ebp 0x100 0x100 %eax 2 300 200 %edx 2 Rtnaddr Old ebp %ebp %esp

  24. Page 4 : Chap. 3 – Stack&Assembly코드의 이해 • pushl %ebp • movl %esp,%ebp • movl8(%ebp),%edx • movl 12(%ebp),%eax • movl %ebp,%esp • movl (%edx),%edx • addl %edx,(%eax) • movl %edx,%eax • popl %ebp • ret 0x300 1 0x10c 2 0x200 %esp 0x108 0x104 0x104 %ebp Old ebp %eax *(0x100+c) 300 300 200 %edx *(0x100+8) 200 Rtnaddr Old ebp %esp

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