这个程序实在看不懂，但愿能有有缘人(80分)解决方案

int mul_s32_s32_s32_sr28(int a, int b){  int result;  unsigned int u32_chi;  unsigned int u32_clo;  mul_wide_s32(a, b, &u32_chi, &u32_clo);  u32_clo = (unsigned int)(u32_chi << 4 | u32_clo >> 28);  result = (unsigned int)u32_clo;  return result;}
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http://www.mathworks.com/matlabcentral/fx_files/27694/1/content/nxtway_gs/models/nxtway_app_fixpt_ert_rtw/html/nxtway_app_fixpt_c

你看看这个 再想想你的程序是用来干嘛的
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C/C++ code
void mul_wide_s32(int in0, int in1, unsigned int *ptrOutBitsHi, unsigned int *ptrOutBitsLo){  unsigned int absIn0;//输入In0的绝对值  unsigned int absIn1;//输入In1的绝对值  unsigned int in0Lo;//输入In0绝对值的低16位  unsigned int in0Hi;//输入In0绝对值的高16位  unsigned int in1Lo;//输入In1绝对值的低16位  unsigned int in1Hi;//输入In1绝对值的高16位  unsigned int productHiLo;//In0的高16位与In的低16位的乘积  unsigned int productLoHi;//In0的低16位与In的高16位的乘积  unsigned int productLoLo;//In0的低16位与In的低16位的乘积        absIn0 = (unsigned int)(in0 < 0 ? -in0 : in0);//对In0取绝对值  absIn1 = (unsigned int)(in1 < 0 ? -in1 : in1);//对In1取绝对值     in0Hi = absIn0>>16;      //get the high 16 bits of int0  in0Lo = absIn0 &65535U;  //get the low 16 bits of int0  in1Hi = absIn1>>16;      //get the high 16 bits of int1  in1Lo = absIn1 &65535U;  //get the low 16 bits of int1                             //as we know that the result should be a 64 bits number   productHiLo = in0Hi*in1Lo; //we get a value whose low 16 bits should be the low 16 bits of final result let us denote the bits by 32-16-0  productLoHi = in0Lo*in1Hi;//let us denote the bits by 48-32-16  productLoLo = in0Lo*in1Lo;//let us denote the bits by 48-32-16                            //you should know 32-16-0 means that the value composes the 32-16-0 bits of the final result     absIn1 = (unsigned int)0; //set absIn1 to zero  in0Lo = (productLoHi << 16) + productLoLo;// as we know that the low 16 bits of final result will come from productLoLo, we need to find is 64-17 bits of the final result, moreover the low 16 bits of productLoHi will compose the 32-17 bits of thefinal result with the high 16 bits of productLoLo, so add themif (in0Lo < productLoLo){absIn1 = (unsigned int)1; //we add the two values, what we get is a number must larger than the original number.If not the case then overflow happens, in other words,the carry must be one. So set absIn1 to unity}absIn0 = in0Lo; in0Lo = (productHiLo << 16) + in0Lo; //the low 16 bits of productHiLo compose the bits 32-17 of final result,so add itif (in0Lo < absIn0){absIn1 = absIn1+(unsigned int)1; //we add the two values, what we get is a number must larger than the original number. If not the case, overflow happens, in other words,the carry must be one, then add the carry to adsIn1}//now let us calculate the 64-33 bits of the final answerabsIn0 = ((in0Hi * in1Hi + absIn1) + (productLoHi >> 16)) + (productHiLo >> 16);//the 64-33 bits of the final answer will composed by in0Hi * in1Hi, high 16 bits of productLoHi, high 16 bits of productHiLo and the carry stored in absIn1//if int0=0, int1=0 or int0 and int1 are both positive numbers we get the final resultif (!((in0 == 0) || (in1 == 0) || ((in0 > 0) == (in1 > 0)))) {    absIn0 = ~absIn0;        in0Lo = ~in0Lo;                           in0Lo = in0Lo + (unsigned int)1; //if result is a negtive number, change result to 2's complement    if (in0Lo == (unsigned int)0) {        absIn0 = absIn0 + (unsigned int)1; //if low 32 bits is zero, then carry must be one we add the carry to the high 32 bits    }}   *ptrOutBitsHi = absIn0; //return the result we get*ptrOutBitsLo = in0Lo;     }